All questions of A Particle Model of Waves for Grade 9 Exam
An echo repeats two syllables. If the velocity of sound is 330 m/s, then the distance of reflecting surface is- a)16.5 m
- b)99 m
- c)66 m
- d)33.0 m
Correct answer is option 'C'. Can you explain this answer?
An echo repeats two syllables. If the velocity of sound is 330 m/s, then the distance of reflecting surface is
a)
16.5 m
b)
99 m
c)
66 m
d)
33.0 m
 | Rajesh Gupta answered |
Let us say that we speak syllables at a rate of 2 to 9 per second. So let us say that a syllable takes a minimum of 0.1 sec for a fast speaker. Let us say that a sound pulse (syllable) is emitted starting at t = 0.
The effect of a syllable lasts on the ear for 0.1 sec. So if any echo reaches the year before t = 0.2 sec., then it is mixed with the direct sound present in the ear and so echo is not properly heard.
In this problem, two syllables are repeated in the echo. That is it took about 2 * 0.2 sec ie., 0.4 seconds for the sound to travel to the reflecting surface and come back to the ear.
The distance of the reflecting surface from the person
= 330 m/s * 0.4 sec / 2
= 66 meters.
A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork,(assuming it is very small when compared to sound) if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s. - a)1 m/s
- b)2 m/s
- c)0.5 m/s
- d)1.5 m/s
Correct answer is option 'C'. Can you explain this answer?
A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork,(assuming it is very small when compared to sound) if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s.
a)
1 m/s
b)
2 m/s
c)
0.5 m/s
d)
1.5 m/s
 | Lavanya Menon answered |
Let the frequencies be f1 and f2
So, f1 = f (vs/vs - v) and f2 = f (vs/vs + v)
ATQ: f1 – f2 = 2 Hz
So, f1 = f (vs/vs - v) and f2 = f (vs/vs + v)
ATQ: f1 – f2 = 2 Hz
At t= 0, a transverse wave pulse travelling in the positive x direction with a speed of 2 m/s in a wire is described by the function y = 6/x2 given that x ≠ 0. Transverse velocity of a particle at x = 2 m and t = 2 s is - a)3 m/s
- b)-3 m/s
- c)8 m/s
- d)-8 m/s
Correct answer is option 'B'. Can you explain this answer?
At t= 0, a transverse wave pulse travelling in the positive x direction with a speed of 2 m/s in a wire is described by the function y = 6/x2 given that x ≠ 0. Transverse velocity of a particle at x = 2 m and t = 2 s is
a)
3 m/s
b)
-3 m/s
c)
8 m/s
d)
-8 m/s
 | Preeti Khanna answered |
Any transverse wave travelling in positive x-direction is of the form y = f (ax - bt) (and y = f (ax + bt) if it is travelling in negative x-direction) and has a wave speed equal to b/a.
Given a = 1 ⇒ b = 2
Given a = 1 ⇒ b = 2
The speed of propagation of a sinusoidal wave is given by V=νλ where- a)ν is the reciprocal of the period and λ is the dispersion
- b)ν is the reciprocal of the period and λ is the wave number
- c)ν is the angular frequency and λ is the wavelength
- d)ν is the reciprocal of the period and λ is the wavelength
Correct answer is option 'D'. Can you explain this answer?
The speed of propagation of a sinusoidal wave is given by V=νλ where
a)
ν is the reciprocal of the period and λ is the dispersion
b)
ν is the reciprocal of the period and λ is the wave number
c)
ν is the angular frequency and λ is the wavelength
d)
ν is the reciprocal of the period and λ is the wavelength
 | Om Desai answered |
For a sinusoidal wave, V = v λ.
V = speed,
v = frequency,
λ = wavelength,
frequency (v) = reciprocal of the time period i.e. v =1/T
V = speed,
v = frequency,
λ = wavelength,
frequency (v) = reciprocal of the time period i.e. v =1/T
A 100 m long rod of density 10.0 x 104 kg/m3 and having Young’s modules Y = 1011 Pa, is clamped at one end. It is hammered at the other free end. The longitudinal pulse goes to right end, gets reflected and again returns to the left end. How much time, the pulse take to go back to initial point.
- a)0.1 sec
- b)0.2 sec
- c)0.3 sec
- d)2 sec
Correct answer is option 'B'. Can you explain this answer?
A 100 m long rod of density 10.0 x 104 kg/m3 and having Young’s modules Y = 1011 Pa, is clamped at one end. It is hammered at the other free end. The longitudinal pulse goes to right end, gets reflected and again returns to the left end. How much time, the pulse take to go back to initial point.
a)
0.1 sec
b)
0.2 sec
c)
0.3 sec
d)
2 sec
| | Rajesh Gupta answered |
The velocity of a wave in any rod is given by,
Note that we should take twice the length of the rod since we want time take by the wave to return back to the initial point.
Note that we should take twice the length of the rod since we want time take by the wave to return back to the initial point.
A piano wire having a diameter of 0.90 mm is replaced by another wire of the same length and material but with a diameter of 0.93 mm. If the tension of the wire is kept the same, then the percentage change in the frequency of the fundamental tone is nearly - a)+3%
- b)+3.3 %
- c)-3.3%
- d)-3%
Correct answer is option 'C'. Can you explain this answer?
A piano wire having a diameter of 0.90 mm is replaced by another wire of the same length and material but with a diameter of 0.93 mm. If the tension of the wire is kept the same, then the percentage change in the frequency of the fundamental tone is nearly
a)
+3%
b)
+3.3 %
c)
-3.3%
d)
-3%
| | Preeti Iyer answered |
Out of all the given quantities only frequency of the wire "f” and radius of the wire "R" changes and the remaining doesn't. Also if a quantity doesn't change (or is a constant) its derivative is zero. Given, initial radius = 0.45 mm and final radius = 0.465 mm so change in radius, ΔR = 0.015 mm
Velocity of sound in air is 300 m/s. Then the distance between two successive nodes of a stationary wave of frequency 1000 Hz is.- a)20cm
- b)15 cm
- c)30cm
- d)10 cm
Correct answer is option 'B'. Can you explain this answer?
Velocity of sound in air is 300 m/s. Then the distance between two successive nodes of a stationary wave of frequency 1000 Hz is.
a)
20cm
b)
15 cm
c)
30cm
d)
10 cm
| | Raghav Bansal answered |
Velocity of sound in air= 300m/s =300×100=30000 cm/s
And frequency = 1000 hz
So, wavelength = Velocity/frequency
= 30000/1000= 30
Distance = wavelength/2
=30/2 = 15
And frequency = 1000 hz
So, wavelength = Velocity/frequency
= 30000/1000= 30
Distance = wavelength/2
=30/2 = 15
A fork of unknown frequency when sounded with another fork of frequency 256 Hz produces 4 beats/sec. The first fork is loaded with wax. It again produces 4 beats/sec. When sounded together with the fork of 256 Hz frequency, then the frequency of first tuning fork is- a)252 Hz
- b)260Hz
- c)None of these
- d)252 or 260 Hz
Correct answer is option 'B'. Can you explain this answer?
A fork of unknown frequency when sounded with another fork of frequency 256 Hz produces 4 beats/sec. The first fork is loaded with wax. It again produces 4 beats/sec. When sounded together with the fork of 256 Hz frequency, then the frequency of first tuning fork is
a)
252 Hz
b)
260Hz
c)
None of these
d)
252 or 260 Hz
| | Raghav Bansal answered |
After waxing the frequency of the tuning fork decreases and so the initial frequency must be higher so as to decrease it below 256 Hz in order to give 4 beats/sec.
In an interference arrangement similar to Young's double-slit experiment, the slits S1 & S2 are illuminated with coherent microwave sources, each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are separated by a distance d = 150.0 m. The intensity I(q) is measured as a function of q, where q is defined as shown. If I0 is the maximum intensity then I(q) for 0 £ q £ 90° is given by 
- a) I(q) =
for q = 30º - b)I(q) =
for q = 90º - c) I(q) = I0 for q = 0º
- d)I(q) is constant for all values of q
Correct answer is option 'A,C'. Can you explain this answer?
In an interference arrangement similar to Young's double-slit experiment, the slits S1 & S2 are illuminated with coherent microwave sources, each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are separated by a distance d = 150.0 m. The intensity I(q) is measured as a function of q, where q is defined as shown. If I0 is the maximum intensity then I(q) for 0 £ q £ 90° is given by
a)
I(q) = for q = 30º
for q = 30º b)
I(q) = for q = 90º
for q = 90º c)
I(q) = I0 for q = 0º
d)
I(q) is constant for all values of q
 | Lohit Matani answered |
I=I0cos2(πdtanθ/ λ)/
= I0cos2(πx150xtanθ/3x(108/106)
=I0cos2(πtanθ/2)
θ=30o
I= I0cos2 ((π/2√3)
I=I0/2
θ=0
I=I0cos2(0) [cos0=1]
I= I0
= I0cos2(πx150xtanθ/3x(108/106)
=I0cos2(πtanθ/2)
θ=30o
I= I0cos2 ((π/2√3)
I=I0/2
θ=0
I=I0cos2(0) [cos0=1]
I= I0
A sinusoidal wave is generated by moving the end of a string up and down, periodically. The generator must apply the energy at maximum rate when the end of the string attached to generator has X and least power when the end of the string attached to generator has Y. The most suitable option which correctly fills blanks X and Y, is - a)Maximum displacement, least acceleration
- b)maximum displacement, maximum acceleration
- c)Least displacement, maximum acceleration
- d)Least displacement, least acceleration
Correct answer is option 'C'. Can you explain this answer?
A sinusoidal wave is generated by moving the end of a string up and down, periodically. The generator must apply the energy at maximum rate when the end of the string attached to generator has X and least power when the end of the string attached to generator has Y. The most suitable option which correctly fills blanks X and Y, is
a)
Maximum displacement, least acceleration
b)
maximum displacement, maximum acceleration
c)
Least displacement, maximum acceleration
d)
Least displacement, least acceleration
 | Juhi Iyer answered |
Power for a travelling wave on a string is given by
For the displacement wave, y = A sin (kx – ωt)
Power delivered is maximum when cos2(kx – ωt) is maximum, which would be the case when sin (kx – ωt) is the least, i.e., displacement is minimum (acceleration is minimum). Power delivered is minimum when cos2(kx – ωt) is minimum, which would be when sin(kx - cos2(kx – ωt)t) is maximum, i.e displacement is maximum(acceleration is maximum).
A beam of light consisting of two wavelengths 6500 Å and 5200 Å is used to obtain interference fringes in Young's double slit experiment. The distance between slits is 2mm and the distance of screen from slits is 120 cm. What is the least distance from central maximum where the bright due to both wavelength coincide ?- a)0.156 cm
- b)0.312 cm
- c)0.078 cm
- d)0.468 cm
Correct answer is option 'A'. Can you explain this answer?
A beam of light consisting of two wavelengths 6500 Å and 5200 Å is used to obtain interference fringes in Young's double slit experiment. The distance between slits is 2mm and the distance of screen from slits is 120 cm. What is the least distance from central maximum where the bright due to both wavelength coincide ?
a)
0.156 cm
b)
0.312 cm
c)
0.078 cm
d)
0.468 cm
 | Gaurav Kumar answered |
If n is least number of fringes of λ1(=6500A∘) which are coincident with (n+1) of smaller wavelength
also β=λD/d
substituting values for wavelength,D and d we get β = 0.039cm
So,
Hence b is the correct answer.
also β=λD/d
substituting values for wavelength,D and d we get β = 0.039cm
So,
Hence b is the correct answer.
A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork,(assuming it is very small when compared to sound) if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s. - a)1 m/s
- b)2 m/s
- c)0.5 m/s
- d)1.5 m/s
Correct answer is option 'C'. Can you explain this answer?
A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork,(assuming it is very small when compared to sound) if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s.
a)
1 m/s
b)
2 m/s
c)
0.5 m/s
d)
1.5 m/s
 | Samridhi Kaur answered |
There are three sources of sound of equal intensities with frequencies 400, 401 and 402 Hz. The number of beats per seconds is- a)3
- b)1.0
- c)0
- d)2
Correct answer is option 'B'. Can you explain this answer?
There are three sources of sound of equal intensities with frequencies 400, 401 and 402 Hz. The number of beats per seconds is
a)
3
b)
1.0
c)
0
d)
2
 | Hansa Sharma answered |
Resultant displacement of the wave by these three wave is
y=asin2π400t+asin2π401t+asin2π402t
y=a(1+2cos2πt)sin2π401t
So the resultant magnitude a(1+2cos2πt) has a maximum when,
cos2πt=1
or, t=0,1,2...
The time interval between two successive maximum is 1 sec.
So beat frequency is 1sec.
y=asin2π400t+asin2π401t+asin2π402t
y=a(1+2cos2πt)sin2π401t
So the resultant magnitude a(1+2cos2πt) has a maximum when,
cos2πt=1
or, t=0,1,2...
The time interval between two successive maximum is 1 sec.
So beat frequency is 1sec.
In a standard YDSE appratus a thin film (m = 1.5, t = 2.1 mm) is placed in front of upper slit. How far above or below the centre point of the screen are two nearest maxima located ? Take D = 1 m, d = 1mm, l = 4500 Å. (Symbols have usual meaning)- a) 1.5 mm (
- b)0.6 mm
- c)0.15 mm
- d)0.3 mm
Correct answer is option 'C,D'. Can you explain this answer?
In a standard YDSE appratus a thin film (m = 1.5, t = 2.1 mm) is placed in front of upper slit. How far above or below the centre point of the screen are two nearest maxima located ? Take D = 1 m, d = 1mm, l = 4500 Å. (Symbols have usual meaning)
a)
1.5 mm (
b)
0.6 mm
c)
0.15 mm
d)
0.3 mm
| | Geetika Shah answered |
A conveyor belt moves to the right with speed v=300 m/min. A pieman puts pies on the belt at a rate of 20 per minute while walking with speed 30 m/min towards a receiver at the other end. The frequency with which they are received by the stationary receiver is : - a)26.67 / minute
- b)30 / minute
- c)22.22/minute
- d)24 / minute
Correct answer is option 'C'. Can you explain this answer?
A conveyor belt moves to the right with speed v=300 m/min. A pieman puts pies on the belt at a rate of 20 per minute while walking with speed 30 m/min towards a receiver at the other end. The frequency with which they are received by the stationary receiver is :
a)
26.67 / minute
b)
30 / minute
c)
22.22/minute
d)
24 / minute
 | Nandini Chakraborty answered |
This problem is a Doppler effect analogy
Where f = 20/min , v= 300 m/min
And vs = 30 m / min
The waves on the surface of water are of two kinds:.- a)capillary waves and gravity waves
- b)capillary waves and sound waves
- c)sound waves and gravity waves
- d)seismic waves and cosmic waves
Correct answer is option 'A'. Can you explain this answer?
The waves on the surface of water are of two kinds:.
a)
capillary waves and gravity waves
b)
capillary waves and sound waves
c)
sound waves and gravity waves
d)
seismic waves and cosmic waves
 | Srishti Roy answered |
Explanation:
As capillary & gravity waves are elastic waves or mechanical waves which require medium for their propagation.
Hence they are using the elastic behaviour of water.
Hence
The waves on the surface of water are of two kinds:
capillary waves and gravity waves
Longitudinal waves cannot be propagated through- a)a liquid
- b)a solid
- c)vacuum
- d)a gas
Correct answer is option 'C'. Can you explain this answer?
Longitudinal waves cannot be propagated through
a)
a liquid
b)
a solid
c)
vacuum
d)
a gas
| | Geetika Shah answered |
Because longitudinal waves are the mechanical waves that need a medium to propagate such as air, gas, solid etc. but these are not available in vacuum, so this wave can't propagate in vacuum.
If a star emitting orange light moves away from the earth, its color will- a)appear yellow
- b)turns gradually blue
- c)remain the same
- d)appear red
Correct answer is option 'D'. Can you explain this answer?
If a star emitting orange light moves away from the earth, its color will
a)
appear yellow
b)
turns gradually blue
c)
remain the same
d)
appear red
 | Anjali Sharma answered |
The faster a star moves towards the earth, the more its light is shifted to higher frequencies. In contrast, if a star is moving away from the earth, its light is shifted to lower frequencies on the color spectrum (towards the orange/red/infrared/microwave/radio end of the spectrum).
A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Distance between the two points having amplitude 2 mm is: - a)1 m
- b)75cm
- c)60cm
- d)50 cm
Correct answer is option 'A'. Can you explain this answer?
A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Distance between the two points having amplitude 2 mm is:
a)
1 m
b)
75cm
c)
60cm
d)
50 cm
 | Jyoti Kumar answered |
**Given:**
- Length of the string = 1.5 m
- Amplitude at the center of the string = 4 mm
- Amplitude at two points = 2 mm
**To find:**
- Distance between the two points having an amplitude of 2 mm
**Solution:**
The fundamental frequency of a string fixed at both ends is given by the formula:
f = (v/2L)
Where,
- f is the frequency of vibration
- v is the velocity of the wave
- L is the length of the string
In the fundamental mode, the wavelength (λ) is double the length of the string. So,
λ = 2L
The velocity of the wave is given by the formula:
v = fλ
Substituting the value of λ, we get:
v = f * 2L
In the fundamental mode, the amplitude is maximum at the center of the string and zero at the ends. So, the amplitude decreases linearly from the center to the ends.
Let's consider the distance (x) from the center of the string to a point where the amplitude is 2 mm.
Using the amplitude formula,
A = (2mm) = (4mm) * (1 - 2x/L)
Simplifying the equation,
1/2 = 1 - 2x/L
2x/L = 1/2
x/L = 1/4
So, the distance between the two points having an amplitude of 2 mm is 1/4 times the length of the string.
Substituting the given length of the string, we get:
x = (1/4) * 1.5m
x = 0.375m
Converting to centimeters,
x = 0.375m * 100cm/m
x = 37.5cm
Therefore, the distance between the two points having an amplitude of 2 mm is 37.5 cm, which is closest to option 'A' (1m).
- Length of the string = 1.5 m
- Amplitude at the center of the string = 4 mm
- Amplitude at two points = 2 mm
**To find:**
- Distance between the two points having an amplitude of 2 mm
**Solution:**
The fundamental frequency of a string fixed at both ends is given by the formula:
f = (v/2L)
Where,
- f is the frequency of vibration
- v is the velocity of the wave
- L is the length of the string
In the fundamental mode, the wavelength (λ) is double the length of the string. So,
λ = 2L
The velocity of the wave is given by the formula:
v = fλ
Substituting the value of λ, we get:
v = f * 2L
In the fundamental mode, the amplitude is maximum at the center of the string and zero at the ends. So, the amplitude decreases linearly from the center to the ends.
Let's consider the distance (x) from the center of the string to a point where the amplitude is 2 mm.
Using the amplitude formula,
A = (2mm) = (4mm) * (1 - 2x/L)
Simplifying the equation,
1/2 = 1 - 2x/L
2x/L = 1/2
x/L = 1/4
So, the distance between the two points having an amplitude of 2 mm is 1/4 times the length of the string.
Substituting the given length of the string, we get:
x = (1/4) * 1.5m
x = 0.375m
Converting to centimeters,
x = 0.375m * 100cm/m
x = 37.5cm
Therefore, the distance between the two points having an amplitude of 2 mm is 37.5 cm, which is closest to option 'A' (1m).
A plane sound wave is travelling in a medium. In a reference frame A fixed to the medium , its equation is y = a cos(ωt − kx ) . With respect to reference frame B, moving with a constant velocity v in the direction of propagation of the wave, equation of the wave will be - a)y = a cos[(ωt + kv) t − kx ]
- b)y = −a cos[(ωt − kv) t − kx ]
- c)y = a cos[(ωt − kv) t − kx ]
- d)y = a cos[(ωt + kv) t + kx ]
Correct answer is option 'C'. Can you explain this answer?
A plane sound wave is travelling in a medium. In a reference frame A fixed to the medium , its equation is y = a cos(ωt − kx ) . With respect to reference frame B, moving with a constant velocity v in the direction of propagation of the wave, equation of the wave will be
a)
y = a cos[(ωt + kv) t − kx ]
b)
y = −a cos[(ωt − kv) t − kx ]
c)
y = a cos[(ωt − kv) t − kx ]
d)
y = a cos[(ωt + kv) t + kx ]
 | Manisha Mehta answered |
Suppose at an instant t, the x – coordinate of a point with reference to moving frame is x0. Since at this moment, origin of moving frame is at distance vt from origin of the fixed reference frame, therefore, putting this value of x in the given equation, we get
y = a cos[ωt − k (vt − x )]
y = a cos[(ωt − kv)t − kx ]
The quantity similar to extension or compression of the spring in sound wave propagation (air) is- a)the change in air density
- b)the change in air composition
- c)the change in air particles
- d)the change in air humidity
Correct answer is option 'A'. Can you explain this answer?
The quantity similar to extension or compression of the spring in sound wave propagation (air) is
a)
the change in air density
b)
the change in air composition
c)
the change in air particles
d)
the change in air humidity
 | Saumya Ahuja answered |
Explanation:As in air wave propagates in the form of compression (increase in density of air) and rarefaction (decrease in density of air).
When sound travels from air to water the quantity that remains unchanged is- a)wavelength
- b)frequency
- c)speed
- d)intensity
Correct answer is option 'B'. Can you explain this answer?
When sound travels from air to water the quantity that remains unchanged is
a)
wavelength
b)
frequency
c)
speed
d)
intensity
| | Ram Hande answered |
Because, frequency of a wave depend on source V1/λ1=V2/λ2 as frequency is constant V1-velocity of sound wave in air V2-velocity of sound wave in water
When we make a mobile telephone call to a friend- a)the friend's mobile receives electromagnetic waves containing your audio
- b)the friend's mobile receives acoustic waves containing your audio
- c)the friend's mobile receives gravity waves containing your audio
- d)the friend's mobile generates possible electrical signals
Correct answer is option 'A'. Can you explain this answer?
When we make a mobile telephone call to a friend
a)
the friend's mobile receives electromagnetic waves containing your audio
b)
the friend's mobile receives acoustic waves containing your audio
c)
the friend's mobile receives gravity waves containing your audio
d)
the friend's mobile generates possible electrical signals
| | Kritika Bajaj answered |
Explanation:Because mobile communication is a space communication and in space communication basically electromagnetic waves are used (as carrier waves as in case of radio communication) because of the modulation ( frequency, amplitude) operations which can be performed on EM waves. Thus when our friend receives the call, he also receives EM waves which is the carrier of our audio signals.
A police car moving at 22 ms-1 , chases a motorcyclist. The police man sounds his horn at 176 Hz, while both of them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcycle, if it is given that the motor cyclist does not observe any beats. (Velocity of sound is 330 ms −1 ) - a)33 ms−1
- b)22 ms−1
- c)11 ms −1
- d)zero
Correct answer is option 'B'. Can you explain this answer?
A police car moving at 22 ms-1 , chases a motorcyclist. The police man sounds his horn at 176 Hz, while both of them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcycle, if it is given that the motor cyclist does not observe any beats. (Velocity of sound is 330 ms −1 )
a)
33 ms−1
b)
22 ms−1
c)
11 ms −1
d)
zero
| | Samridhi Kaur answered |
Frequency of the police horn
Heard by the motor cyclist is
Frequency of the stationary Siren heard by the motor cyclist is
This gives v = 22ms −1 .
The velocity of sound in gas in which two waves of wavelength 1.0 m and 0.01 m produces 4 beats/sec. is- a)360 m/s
- b)1010 m/s
- c)404 m/s
- d)440 m/s
Correct answer is option 'C'. Can you explain this answer?
The velocity of sound in gas in which two waves of wavelength 1.0 m and 0.01 m produces 4 beats/sec. is
a)
360 m/s
b)
1010 m/s
c)
404 m/s
d)
440 m/s
| | Ruchi Dasgupta answered |
Beats = |f1 – f2| where f1 and f2 are the frequencies
Now, f1 = v/λ1 and f2 = v/λ2
So, 4 = |v (1/1- 1/0.01)|
v = 0.0404 m/s
Now, f1 = v/λ1 and f2 = v/λ2
So, 4 = |v (1/1- 1/0.01)|
v = 0.0404 m/s
Speed of sound in air is 350 m/s. An engine blows a whistle of frequency of 1200 Hz, it is approaching the observer with velocity 50 m/s. The apparent frequency as heard by the observer is- a)300 Hz
- b)1400 Hz
- c)1600 Hz
- d)1050 Hz
Correct answer is option 'B'. Can you explain this answer?
Speed of sound in air is 350 m/s. An engine blows a whistle of frequency of 1200 Hz, it is approaching the observer with velocity 50 m/s. The apparent frequency as heard by the observer is
a)
300 Hz
b)
1400 Hz
c)
1600 Hz
d)
1050 Hz
 | Anonymous answered |
An organ pipe closed at one end has a fundamental frequency of 256 Hz. If the speed of sound in air is 340 m/s, the length of the pipe is approximately:- a)0.33 m
- b)0.66 m
- c)1.32 m
- d)2.64 m
Correct answer is option 'A'. Can you explain this answer?
a)
0.33 m
b)
0.66 m
c)
1.32 m
d)
2.64 m
 | Aryan Dasgupta answered |
Understanding the Fundamental Frequency
The fundamental frequency of an organ pipe closed at one end is determined by the length of the pipe and the speed of sound in air. For a pipe closed at one end:
- The fundamental frequency (f) is given by the formula:
f = v / 4L
Where:
- f = fundamental frequency
- v = speed of sound in air
- L = length of the pipe
Given Data
- Fundamental frequency (f) = 256 Hz
- Speed of sound (v) = 340 m/s
Calculating the Length of the Pipe
To find the length (L) of the pipe, we can rearrange the formula:
L = v / (4f)
Now, substituting the values:
- v = 340 m/s
- f = 256 Hz
L = 340 / (4 * 256)
L = 340 / 1024
L ≈ 0.332 m
Conclusion
The length of the pipe is approximately 0.33 m, which corresponds to option a).
This calculation confirms that the fundamental frequency of an organ pipe closed at one end can be effectively used to determine its length using the relationship between frequency, speed of sound, and pipe length.
The fundamental frequency of an organ pipe closed at one end is determined by the length of the pipe and the speed of sound in air. For a pipe closed at one end:
- The fundamental frequency (f) is given by the formula:
f = v / 4L
Where:
- f = fundamental frequency
- v = speed of sound in air
- L = length of the pipe
Given Data
- Fundamental frequency (f) = 256 Hz
- Speed of sound (v) = 340 m/s
Calculating the Length of the Pipe
To find the length (L) of the pipe, we can rearrange the formula:
L = v / (4f)
Now, substituting the values:
- v = 340 m/s
- f = 256 Hz
L = 340 / (4 * 256)
L = 340 / 1024
L ≈ 0.332 m
Conclusion
The length of the pipe is approximately 0.33 m, which corresponds to option a).
This calculation confirms that the fundamental frequency of an organ pipe closed at one end can be effectively used to determine its length using the relationship between frequency, speed of sound, and pipe length.
A stretched string of length 1m fixed at both ends, having a mass 5×10−4kg is under a tension of 20N. It is plucked at a point situated at 25 cm from one end. The stretched string could vibrate with a frequency of - a)512 Hz
- b)100 Hz
- c)200 Hz
- d)256 Hz
Correct answer is option 'C'. Can you explain this answer?
A stretched string of length 1m fixed at both ends, having a mass 5×10−4kg is under a tension of 20N. It is plucked at a point situated at 25 cm from one end. The stretched string could vibrate with a frequency of
a)
512 Hz
b)
100 Hz
c)
200 Hz
d)
256 Hz
 | Shruti Ahuja answered |
At 25 cm, there will be antinode. So wire will vibrate in two loops
Two waves of wavelength 1m & 1.01 m produce 10 beats in 3 sec. The velocity of sound in a gas is about- a)300 m/s
- b)337 m/sec
- c)33 m/s
- d)1120 m/sec
Correct answer is option 'B'. Can you explain this answer?
Two waves of wavelength 1m & 1.01 m produce 10 beats in 3 sec. The velocity of sound in a gas is about
a)
300 m/s
b)
337 m/sec
c)
33 m/s
d)
1120 m/sec
 | Mira Sharma answered |
As we know wavelength= spped/frequency
Wavelength 1= 1m
Wavelength 2= 1.01m
=> 1=v/f1
f1= v
1.01=v/f2
f2= v/1.01
Now beat is 19 beats /3 sec
f1 - f2= 10/3
Solving equations we get
v=336.6m/s
Two sinusoidal waves on the same string exhibit interference, adding or cancelling according to the principle of superposition. If the two are travelling in the same direction and have the same amplitude a and frequency but differ in phase by a phase constant φφ, the result is a single wave with the same frequency ω:If φ = 0 or an integral multiple of 2 π- a)the waves are exactly in phase and the interference is constructive.
- b)the waves are exactly out of phase and the interference is desstructive.
- c)the waves are exactly in phase and the interference is destructive.
- d)the waves are exactly out of phase and the interference is constructive.
Correct answer is option 'A'. Can you explain this answer?
Two sinusoidal waves on the same string exhibit interference, adding or cancelling according to the principle of superposition. If the two are travelling in the same direction and have the same amplitude a and frequency but differ in phase by a phase constant φφ, the result is a single wave with the same frequency ω:If φ = 0 or an integral multiple of 2 π
a)
the waves are exactly in phase and the interference is constructive.
b)
the waves are exactly out of phase and the interference is desstructive.
c)
the waves are exactly in phase and the interference is destructive.
d)
the waves are exactly out of phase and the interference is constructive.
 | Jhanvi Chakraborty answered |
Let the waves be
y1 = Asin(wt)
y2 = Asin(wt + φ)
If φ = 0 or 2nπ then the waves are exactly in phase and the interference is constructive.
Hence A is the correct answer.
y1 = Asin(wt)
y2 = Asin(wt + φ)
If φ = 0 or 2nπ then the waves are exactly in phase and the interference is constructive.
Hence A is the correct answer.
A chord attached about an end to a vibrating fork divides it into 6 loops. When its tension is 36N. The tension at which it will vibrate in 4 loops when attached to same tuning fork is: - a)24 N
- b)36 N
- c)64 N
- d)81 N
Correct answer is option 'D'. Can you explain this answer?
A chord attached about an end to a vibrating fork divides it into 6 loops. When its tension is 36N. The tension at which it will vibrate in 4 loops when attached to same tuning fork is:
a)
24 N
b)
36 N
c)
64 N
d)
81 N
 | Milan Datta answered |
Given information:
- A chord attached about an end to a vibrating fork divides it into 6 loops.
- The tension in the chord is 36N.
To find:
The tension at which the chord will vibrate in 4 loops when attached to the same tuning fork.
Explanation:
1. Relationship between Tension and Number of Loops:
The number of loops formed by a vibrating string or chord is inversely proportional to the tension in the string. This means that as the tension increases, the number of loops decreases, and vice versa.
2. Applying the Relationship:
In the given scenario, the chord attached to the vibrating fork forms 6 loops when the tension is 36N.
We can use the relationship mentioned above to determine the tension required for 4 loops.
Let's assume the tension required for 4 loops is T.
Since the relationship is inversely proportional, we can set up the following equation:
Tension1 * Number of Loops1 = Tension2 * Number of Loops2
Substituting the given values:
36N * 6 loops = T * 4 loops
Simplifying the equation:
216N = 4T
Dividing both sides by 4:
54N = T
Therefore, the tension required for the chord to vibrate in 4 loops when attached to the same tuning fork is 54N.
However, none of the given options match this answer. Option D is not the correct answer based on the information provided.
Please verify the options or provide additional information if available.
- A chord attached about an end to a vibrating fork divides it into 6 loops.
- The tension in the chord is 36N.
To find:
The tension at which the chord will vibrate in 4 loops when attached to the same tuning fork.
Explanation:
1. Relationship between Tension and Number of Loops:
The number of loops formed by a vibrating string or chord is inversely proportional to the tension in the string. This means that as the tension increases, the number of loops decreases, and vice versa.
2. Applying the Relationship:
In the given scenario, the chord attached to the vibrating fork forms 6 loops when the tension is 36N.
We can use the relationship mentioned above to determine the tension required for 4 loops.
Let's assume the tension required for 4 loops is T.
Since the relationship is inversely proportional, we can set up the following equation:
Tension1 * Number of Loops1 = Tension2 * Number of Loops2
Substituting the given values:
36N * 6 loops = T * 4 loops
Simplifying the equation:
216N = 4T
Dividing both sides by 4:
54N = T
Therefore, the tension required for the chord to vibrate in 4 loops when attached to the same tuning fork is 54N.
However, none of the given options match this answer. Option D is not the correct answer based on the information provided.
Please verify the options or provide additional information if available.
In wave propagation- a)there is no flow of matter and there is no movement of disturbance
- b)there is flow of matter and there is no movement of disturbance
- c)there is flow of matter but there is movement of disturbance
- d)there is no flow of matter but there is movement of disturbance
Correct answer is option 'D'. Can you explain this answer?
In wave propagation
a)
there is no flow of matter and there is no movement of disturbance
b)
there is flow of matter and there is no movement of disturbance
c)
there is flow of matter but there is movement of disturbance
d)
there is no flow of matter but there is movement of disturbance
 | Shanaya Tiwari answered |
Wave Propagation
Wave propagation refers to the transmission of energy through a medium via disturbances in the medium. Waves can be classified into mechanical waves (require a medium to propagate) and electromagnetic waves (can propagate through a vacuum).
Flow of Matter and Movement of Disturbance
- In wave propagation, there is no flow of matter. This means that the particles of the medium do not move along with the wave. Instead, they oscillate about their equilibrium positions.
- However, there is movement of disturbance. The disturbance created by the wave travels through the medium, causing the particles to oscillate but not to flow in the direction of the wave.
Characteristics of Wave Propagation
- Waves transfer energy from one point to another without the physical transfer of matter.
- The disturbance created by a wave propagates through the medium by causing particles in the medium to oscillate.
- Different types of waves (such as sound waves, water waves, and electromagnetic waves) exhibit wave propagation characteristics.
In conclusion, in wave propagation, there is no flow of matter but there is movement of disturbance. This movement of disturbance allows energy to be transmitted through the medium without the physical displacement of matter.
Wave propagation refers to the transmission of energy through a medium via disturbances in the medium. Waves can be classified into mechanical waves (require a medium to propagate) and electromagnetic waves (can propagate through a vacuum).
Flow of Matter and Movement of Disturbance
- In wave propagation, there is no flow of matter. This means that the particles of the medium do not move along with the wave. Instead, they oscillate about their equilibrium positions.
- However, there is movement of disturbance. The disturbance created by the wave travels through the medium, causing the particles to oscillate but not to flow in the direction of the wave.
Characteristics of Wave Propagation
- Waves transfer energy from one point to another without the physical transfer of matter.
- The disturbance created by a wave propagates through the medium by causing particles in the medium to oscillate.
- Different types of waves (such as sound waves, water waves, and electromagnetic waves) exhibit wave propagation characteristics.
In conclusion, in wave propagation, there is no flow of matter but there is movement of disturbance. This movement of disturbance allows energy to be transmitted through the medium without the physical displacement of matter.
The figure shows at time t = 0, a rectangular and triangular pulse on a uniform wire are approaching each other. The pulse speed is 0.5 cm/s. The resultant pulse at t = 2 second
is 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The figure shows at time t = 0, a rectangular and triangular pulse on a uniform wire are approaching each other. The pulse speed is 0.5 cm/s. The resultant pulse at t = 2 second
is
is
a)
b)
c)
d)
 | Vaishnavi Bajaj answered |
At t=2 second, the position of both pulses are separately given by fig(a) and fig(b); he superposition of both pulses is given by fig (c)
Equation of a stationary and a travelling waves are as follows y1 = a sin kx cos ωt and y2 = a sin ( ωt − kx ) . The phase difference between two points 
in the standing wave ( y1 ) and is φ2 in travelling wave (y2) , then ratio 
- a)1
- b)5/6
- c)3/4
- d)6/7
Correct answer is option 'D'. Can you explain this answer?
Equation of a stationary and a travelling waves are as follows y1 = a sin kx cos ωt and y2 = a sin ( ωt − kx ) . The phase difference between two points in the standing wave ( y1 ) and is φ2 in travelling wave (y2) , then ratio
in the standing wave ( y1 ) and is φ2 in travelling wave (y2) , then ratio a)
1
b)
5/6
c)
3/4
d)
6/7
| | Juhi Iyer answered |
x1 and x2 are in successive loops of std. waves.
Two sound sources each emitting waves of wavelength λ are fixed a given distance apart and an observer moves from one source to another with velocity u. Then number of beats heard by him - a) 2u/λ
- b)u/λ
- c)√uλ
- d)u/2λ
Correct answer is option 'A'. Can you explain this answer?
Two sound sources each emitting waves of wavelength λ are fixed a given distance apart and an observer moves from one source to another with velocity u. Then number of beats heard by him
a)
2u/λ
b)
u/λ
c)
√uλ
d)
u/2λ
 | Anshu Joshi answered |
Number of extra waves received from S2 per second = +u/λ
Number of lesser waves received from S1 per second = − u /λ
Beat frequency = u/λ -(- u/λ) = 2u/λ
Number of lesser waves received from S1 per second = − u /λ
Beat frequency = u/λ -(- u/λ) = 2u/λ
In a stationary wave that forms as a result of reflection of waves from an obstacle the ratio of the amplitude at an antinode to the amplitude at nodes is n. The fraction of energy reflected is: - a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
In a stationary wave that forms as a result of reflection of waves from an obstacle the ratio of the amplitude at an antinode to the amplitude at nodes is n. The fraction of energy reflected is:
a)
b)
c)
d)
 | Nabanita Deshpande answered |
Let ai and ar to be the amplitude of incident and reflected wave. Then 
A sound wave of frequency 440 Hz is passing through air. An O2 molecule (mass = 5.3 × 10-26 kg) is set in oscillation with an amplitude of 10-6 m. Its speed at the mean position of oscillation is : - a) 2.76 ×10−3 m / s
- b)17.0 ×10−5 m / s
- c)1.70 ×10−5 m / s
- d)2.77 ×10−5 m / s
Correct answer is option 'A'. Can you explain this answer?
A sound wave of frequency 440 Hz is passing through air. An O2 molecule (mass = 5.3 × 10-26 kg) is set in oscillation with an amplitude of 10-6 m. Its speed at the mean position of oscillation is :
a)
2.76 ×10−3 m / s
b)
17.0 ×10−5 m / s
c)
1.70 ×10−5 m / s
d)
2.77 ×10−5 m / s
| | Hansa Sharma answered |
vmax=ωnA
=(2πf)A
=(2π) =(2π)(440)(10−6)
=2.76×10−3 m/s
=(2πf)A
=(2π) =(2π)(440)(10−6)
=2.76×10−3 m/s
Which one is not produced by sound waves in air?- a)Diffraction
- b)Reflection
- c)Refraction
- d)Polarization
Correct answer is option 'D'. Can you explain this answer?
Which one is not produced by sound waves in air?
a)
Diffraction
b)
Reflection
c)
Refraction
d)
Polarization
| | Rohan Singh answered |
Polarisation is a phenomenon of converting unpolarized light to polarized. It is not done by sound waves.
A whistle producing sound waves of frequencies 9,500 Hz and above is approaching a stationary person with speed u ms-1. The velocity of sound in air is 300 ms-1 . If the person can hear frequencies upto a maximum of 10,000 Hz, The maximum value of u upto which he can hear the whistle is - a)15 √2 ms −1
- b)15 / √2 ms−1
- c)15ms−1
- d)30ms−1
Correct answer is option 'C'. Can you explain this answer?
A whistle producing sound waves of frequencies 9,500 Hz and above is approaching a stationary person with speed u ms-1. The velocity of sound in air is 300 ms-1 . If the person can hear frequencies upto a maximum of 10,000 Hz, The maximum value of u upto which he can hear the whistle is
a)
15 √2 ms −1
b)
15 / √2 ms−1
c)
15ms−1
d)
30ms−1
 | Advait Chakraborty answered |
Travelling or progressive wave- a)does not move from one point of the medium to all others
- b)does not move from one point of the medium to another
- c)does not move from one point of the medium to any other
- d)travels from one point of the medium to another
Correct answer is option 'D'. Can you explain this answer?
Travelling or progressive wave
a)
does not move from one point of the medium to all others
b)
does not move from one point of the medium to another
c)
does not move from one point of the medium to any other
d)
travels from one point of the medium to another
| | Kritika Bajaj answered |
Explanation:As progressive wave means a wave propagating in some onward direction in the medium
in the same medium transverse and longitudinal waves- a)travel changing longitudinal wave to transverse wave
- b)travel with different speeds
- c)travel changing transverse wave to longitudinal wave
- d)travel with same speeds
Correct answer is option 'B'. Can you explain this answer?
in the same medium transverse and longitudinal waves
a)
travel changing longitudinal wave to transverse wave
b)
travel with different speeds
c)
travel changing transverse wave to longitudinal wave
d)
travel with same speeds
 | Rithika Khanna answered |
Explanation:
As speed of transverse & longitudinal waves depend on different modulus of elasticity (Young's modulus, Bulk modulus, Modulus of Rigidity) of the medium.
So in the same medium transverse and longitudinal waves
travel with different speeds
Electromagnetic waves are different from sound waves in that- a)they need no medium and are transverse
- b)they need medium and are longitudinal
- c)they need medium and are transverse
- d)they need no medium and are longitudinal
Correct answer is option 'A'. Can you explain this answer?
Electromagnetic waves are different from sound waves in that
a)
they need no medium and are transverse
b)
they need medium and are longitudinal
c)
they need medium and are transverse
d)
they need no medium and are longitudinal
| | Rithika Khanna answered |
Explanation:Electromagnetic waves are transverse waves, they move perpendicular to the direction of propagation of wave ( the direction in which energy is transferred) and EM waves( Electromagnetic waves) can travel in vacuum, thus doesn't require any medium also.
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