Test: 35 Year JEE Previous Year Questions: Waves
32 Questions MCQ Test Physics For JEE | Test: 35 Year JEE Previous Year Questions: Waves
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Waves y1 = A cos(0.5πx - 100πt) and y2 = A cos(0.46πx - 92πt) are travelling along x-axis. (Here x is in m and t is in second)
Q. Find the number of times intensity is maximum in time interval
The equations are y1 = A cos (0.5 πx – 100 πt) and y2 = A cos (0.46 πx – 92 πt) represents two progressive wave travelling in the same direction with slight difference in the frequency. This will give the phenomenon of beats.
Comparing it with the equation
Wave velocity = λ1f1 = 200 m/s [Alternatively use v = ω/K]
ω2 = 92 π ⇒ 2πf2 = 92 π ⇒ f2 = 46 Hz Therefore beat frequency = f1 – f2 = 4 Hz and
NOTE : Wave velocity is same because it depends on the medium in which the wave is travelling.
Now, at x = 0, y1 + y2 = (A cos 10 πt) + (A cos 92 πt) = 0
⇒ cos 100 πt = – cos 92 πt = cos (– 92 πt)
⇒ net amplitude is zero for n = 96 times (the nearest answer).
Waves y1 = A cos(0.5πx - 100πt) and y2 = A cos(0.46πx - 92πt) are travelling along x-axis. (Here x is in m and t is in second)
Q. The wave velocity of louder sound is
The equations are y1 = A cos (0.5 πx – 100 πt) and y2 = A cos (0.46 πx – 92 πt) represents two progressive wave travelling in the same direction with slight difference in the frequency. This will give the phenomenon of beats.
Comparing it with the equation
Wave velocity = λ1f1 = 200 m/s [Alternatively use v = ω/K]
ω2 = 92 π ⇒ 2πf2 = 92 π ⇒ f2 = 46 Hz Therefore beat frequency = f1 – f2 = 4 Hz and
NOTE : Wave velocity is same because it depends on the medium in which the wave is travelling.
Now, at x = 0, y1 + y2 = (A cos 10 πt) + (A cos 92 πt) = 0
⇒ cos 100 πt = – cos 92 πt = cos (– 92 πt)
⇒ net amplitude is zero for n = 96 times (the nearest answer).
Waves y1 = A cos(0.5πx - 100πt) and y2 = A cos(0.46πx - 92πt) are travelling along x-axis. (Here x is in m and t is in second)
Q. The number of times y1 + y2 = 0 at x = 0 in 1 sec is
The equations are y1 = A cos (0.5 πx – 100 πt) and y2 = A cos (0.46 πx – 92 πt) represents two progressive wave travelling in the same direction with slight difference in the frequency. This will give the phenomenon of beats.
Comparing it with the equation
Wave velocity = λ1f1 = 200 m/s [Alternatively use v = ω/K]
ω2 = 92 π ⇒ 2πf2 = 92 π ⇒ f2 = 46 Hz Therefore beat frequency = f1 – f2 = 4 Hz and
NOTE : Wave velocity is same because it depends on the medium in which the wave is travelling.
Now, at x = 0, y1 + y2 = (A cos 10 πt) + (A cos 92 πt) = 0
⇒ cos 100 πt = – cos 92 πt = cos (– 92 πt)
⇒ net amplitude is zero for n = 96 times (the nearest answer).
Two trains A and B moving with speeds 20 m/s and 30 m/s respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engine of train A blows a long whistle.
Assume that the sound of the whistle is composed of components varying in frequency from f1 = 800 Hz to f2 = 1120 Hz, as shown in the figure. The spread in the frequency (highest frequency – lowest frequency) is thus 320 Hz. The speed of sound in still air is 340 m/s.
Q. The speed of sound of the whistle is
The speed of sound depends on the frame of reference of the observer.
Two trains A and B moving with speeds 20 m/s and 30 m/s respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engine of train A blows a long whistle.
Assume that the sound of the whistle is composed of components varying in frequency from f1 = 800 Hz to f2 = 1120 Hz, as shown in the figure. The spread in the frequency (highest frequency – lowest frequency) is thus 320 Hz. The speed of sound in still air is 340 m/s.
Q. The distribution of the sound intensity of the whistle as observed by the passengers in train A is best represented by
Since all the passengers in train A are moving with a velocity of 20 m/s therefore the distribution of sound intensity of the whistle by the passengers in train A is uniform.
Two trains A and B moving with speeds 20 m/s and 30 m/s respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engine of train A blows a long whistle.
Assume that the sound of the whistle is composed of components varying in frequency from f1 = 800 Hz to f2 = 1120 Hz, as shown in the figure. The spread in the frequency (highest frequency – lowest frequency) is thus 320 Hz. The speed of sound in still air is 340 m/s.
Q. The spread of frequency as observed by the passengers in train B is
Length of a string tied to two rigid supports is 40 cm. Maximum length (wavelength in cm) of a stationary wave produced on it is
This will happen for fundamental mode of vibration as shown in the figure. S1 and S2 are rigid support
Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube A and B is
KEY CONCEPT : The fundamental frequency for closed organ pipe is given by 
and
For open organ pipe is given by 
A tuning fork arrangement (pair) produces 4 beats/sec with one fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/sec. The frequency of the unknown fork is
A tuning fork produces 4 beats/sec with another tuning fork of frequency 288 cps. From this information we can conclude that the frequency of unknown fork is 288 + 4 cps or 288 – 4 cps i.e. 292 cps or 284 cps. When a little wax is placed on the unknown fork, it produces 2 beats/sec. When a little wax is placed on the unknown fork, its frequency decreases and simultaneously the beat frequency decreases confirming that the frequency of the unknown fork is 292 cps.
A wave y = a sin(ωt–kx) on a string meets with another wave producing a node at x = 0. Then the equation of the unknown wave is
To form a node there should be superposition of this wave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of π. The equation of the reflected wave will be
y = a sin (ωt + kx + π)
⇒ y = – a sin (ωt + kx)
When temperature increases, the frequency of a tuning fork
KEY CONCEPT : The frequency of a tuning fork is given by the expression
As temperature increases, ℓ increases and therefore f decreases.
The displacement y of a wave travelling in the x -direction is given by 
where x is expressed in metres and t in seconds. The speed of the wave - motion, in ms-1 , is
But y = A sin (ωt - kx + φ)
On comparing we get ω = 600; k = 2
A metal wire of linear mass density of 9.8 g/m is stretched with a tension of 10 kg-wt between two rigid supports 1 metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency n. The frequency n of the alternating source is
KEY CONCEPT : For a string vibrating between two rigid support, the fundamental frequency is given by
As the string is vibrating in resonance to a.c of frequency n, therefore both the frequencies are same.
A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
A tuning fork of frequency 256 Hz makes 5 beats/ second with the vibrating string of a piano. Therefore the frequency of the vibrating string of piano is (256 ± 5) Hz ie either 261Hz or 251 Hz. When the tension in the piano string increases, its frequency will increases. Now since the beat frequency decreases, we can conclude that the frequency of piano string is 251Hz
The displacement y of a particle in a medium can be expressed as, 
where t is in second and x in meter. The speed of the wave is
From equation given,
When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?
No. of beats heard when fork 2 is sounded with fork 1 = Δn = 4
Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from 4 to 6 in this case) then the frequency of the unknown fork 2 is given by,
n = n0 - Δn = 200 – 4 = 196 Hz
An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound.What is the percentage increase in the apparent frequency ?
A whistle producing sound waves of frequencies 9500 HZ and above is approaching a stationary person with speed v 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 v upto which he can hear whistle is
cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is
Lowest resonant frequency is when n = 1
Therefore lowest resonant frequency = 105 Hz.
A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of
⇒ Intensity decreases by a factor 100.
While measuring the speed of sound by perfor ming a resonance column experiment, a student gets the first resonance condition at a column length of 18 cm during winter. Repeating the same experiment during summer, she measures the column length to be x cm for the second resonance. Then
For first resonant length 
For second resonant length
v' > v because velocity of light is greater in summer as compared to winter 
∴ x > 54 cm
A wave travelling along the x-axis is described by the equation y(x, t) = 0.005 cos (αx –βt). If the wavelength and the time period of the wave are 0.08 m and 2.0s, respectively, then α and β in appropriate units are
y(x, t) = 0.005 cos (αx -βt) (Given)
Comparing it with the standard equation of wave y(x, t) = a cos (kx -ωt) we get
Three sound waves of equal amplitudes have frequencies (v –1), v, (v + 1). They superpose to give beats. The number of beats produced per second will be :
Maximum number of beats = (v + 1) – (v – 1) = 2
A motor cycle starts from rest and accelerates along a straight path at 2m/s2. At the starting point of the motor cycle there is a stationary electric siren. How far has the motor cycle gone when the driver hears the frequency of the siren at 94% of its value when the motor cycle was at rest? (Speed of sound = 330 ms–1)
According to Doppler’s effect
The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by
The tension in the string is
∴ velocity, v = nλ = 25 × 0.5 m/s = 12.5 m/s
Velocity on a string is given by
The transverse displacement y (x, t) of a wave on a string is given by 
. This represents a:
A cylindrical tube, open at both ends, has a fundamental frequency, f, in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now :
The fundamental frequency of open tube
That of closed pipe
According to the problem 
From equations (i) and (iii)
A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and 2.2 × 1011 N/m2 respectively ?
Fundamental frequency,
Putting the value of 
A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.
Length of pipe = 85 cm = 0.85m Frequency of oscillations of air column in closed organ pipe is given by,
A train is moving on a straight track with speed 20 ms–1. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms–1) close to :
A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the supports is :
(take g = 10 ms–2)
We know that velocity in string is given by
From (a) and (b)
A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now :
The fundamental frequency in case (a) is 
The fundamental frequency in case (b) is
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