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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?

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 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?

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?

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

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?

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). 

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?

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.

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?

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?

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?

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?

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).

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?

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 ]

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?

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?

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.

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?

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.

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?

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

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?

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.

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?

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.

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?

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?

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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