## The two waves are represented by : y1 = 10-6 sin(100t+x/50+0.5)m and y2 = 10-2 cos(100t+x/50)m

Q: The two waves are represented by :

y1 = 10-6 sin(100t + x/50 + 0.5) m

and y2 = 10-2 cos(100t + x/50) m

where x is in metres and t in seconds. The phase difference between the waves is approximately:

Ans: (a)

Solution : y1 = 10-6 sin(100t + x/50 + 0.5)m

and y2 = 10-2 sin(100t + x/50 + π/2)m

y2 = 10-2 sin(100t + x/50 + 1.57 )m

Phase difference = (100 t + x/50 + 1.57 ) – (100 t + x/50 + 0.5 )

## In Resonance tube experiment, if 400 Hz tuning fork is used, the first resonance occurs when length of air column …

Q: In Resonance tube experiment, if 400 Hz tuning fork is used, the first resonance occurs when length of air column in the tube is 19 cm. If the 400 Hz. tuning fork is replaced by 1600 Hz tuning fork then to get resonance, the water level in the tube should be further lowered by (take end correction = 1 cm)

(A) 5 cm

(B) 10 cm

(C) 15 cm

(D) 20 cm

Ans: (A),(C)

Solution: For first resonance with 400 Hz tuning fork

$\displaystyle L_{eq} = \frac{v}{4 f_0 }$

$\displaystyle L_{eq} = \frac{v}{4 (400) } = (19+1) = 20 cm$ $\displaystyle \frac{v}{4 f_0 } = \frac{v}{4 (1600) } = \frac{20}{4} = 5 cm$

For Resonance

$\displaystyle L_{eq} = \frac{v}{4 f_0 } , \frac{3 v}{4 f_0 } , \frac{5 v}{4 f_0 }, \frac{7 v}{4 f_0 }….$

1 cm + L = 5 cm , 15 cm , 25 cm , 35 cm , 45 cm …..

L = 4 cm , 14 cm , 24 cm , 34 cm , 44 cm …..

water level should be further lowered by
24 – 19 = 5 cm

⇒ 34 – 19 = 15 cm

## The source (S) of sound is moving constant velocity v0 as shown in diagram. An observer O listens to the sound ….

Q: The source (S) of sound is moving constant velocity v0 as shown in diagram. An observer O listens to the sound emitted by the source. The observed frequency of the sound (A) continuously decreases

(B) continuously increases

(C) first decreases then increases

(D) first increases then decreases

Ans: (A)

Solution: From figure , the velocity of approach (Vcosθ) decrease as the source comes closer (as θ increases).And the velocity of separation also increases as φ will decrease. Hence the frequency of sound as heard by the observed decreases continuously . ## A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode….

Q: 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) 75 cm

(C) 60 cm

(D) 50 cm

Ans: (A)

Solution: λ = 2 l = 3 m

Equation of standing wave

y = 2A sin kx cos ωty = A as amplitude is 2A

A = 2A sin kx

$sin k x = \frac{1}{2}$

$k x = \frac{\pi}{6}$

$\frac{2\pi}{\lambda} x = \frac{\pi}{6}$

$\frac{2\pi}{3} x = \frac{\pi}{6}$

x1 = 1/4 m

And , $k x = \frac{5 \pi}{6}$

$\frac{2\pi}{\lambda} x = \frac{5 \pi}{6}$

x2 = 1.25 m

x2 – x1 = 1 m

## A sinusoidal wave is propagating in negative x-direction in a string stretched along x-axis.A particle of string …

Q: COMPREHENSION:
A sinusoidal wave is propagating in negative x-direction in a string stretched along x-axis. A particle of string
at x = 2 m is found at its mean position and it is moving in positive y direction at t = 1 sec. If the amplitude
of the wave, the wavelength and the angular frequency of the wave are 0.1 meter, π/4 meter and 4π rad/sec
respectively.

1 .The equation of the wave is

(A) y = 0.1 sin (4πt – 1)+ 8(x – 2))

(B) y = 0.1 sin (t-1)- (x – 2))

(C) y = 0.1 sin (4π(t -1)-8(x – 2))

(D) none of these

2. The speed of particle at x = 2 m and t = 1 sec is

(A) 0.2π m/s

(B) 0.6π m/s

(C) 0.4π m/s

(D) 0

3. The instantaneous power transfer through x = 2 m and t = 1.125 sec, is
(A) 10 J/s

(B) 4π/3 J/s

(C) 2π/3 J/s

(D) zero

Ans: 1.(A) ; 2. (C) ; 3.(D)

Solution: 1 . The equation of wave moving in negative x-direction, assuming origin of position at x = 2 and origin of time
(i.e. initial time) at t = 1 sec. y = 0.1 sin (4πt + 8x)
Shifting the origin of position to left by 2m, that is, to
x = 0. Also shifting the origin of time backwards by 1 sec,
that is to t = 0 sec. y = 0.1 sin [(4πt + 8(x – 2)]

2. As given the particle at x = 2 is at mean position at t = 1 sec.  its velocity v = ωA = 4π × 0.1 = 0.4 π m/s.

3. Time period of oscillation T = 2π/ω sec. Hence at t = 1.125 sec, that is, at T/4 seconds after t = 1 second, the particle is at rest at extreme position. Hence instantaneous power at x = 2 at t = 1.125 sec is zero.

## A car moves towards a hill with speed vc. It blows a horn of frequency f which is heard by an observer ….

Q: A car moves towards a hill with speed vc. It blows a horn of frequency f which is heard by an observer following
the car with speed v0 . The speed of sound in air

(A) the wavelength of sound reaching the hill is $\frac{v}{f}$

(B) the wavelength of sound reaching the hill is $\frac{v -v_c}{f}$

(C) The wavelength of sound of horn directly reaching the observer is $\frac{v + v_c}{f}$

(D) the beat frequency observed by the observer is $\frac{2v_c(v+ v_o)f}{v^2 – v_c^2}$

Ans: (D)

Solution: Frequency of echo $= (\frac{v}{v+v_c})f$

Frequency of echo of horn as heard by observer $= (\frac{v}{v-v_c})f (\frac{v+v_o}{v})$

Frequency of Beats : $= (v+v_o)f(\frac{1}{v-v_c} – \frac{1}{v+v_c})$

$= \frac{2v_c(v+ v_o)f}{v^2 – v_c^2}$

## COMPREHENSION : Figure shows a clamped metal string of length 30 cm and linear mass density 0.1 kg/m….

COMPREHENSION :
Figure shows a clamped metal string of length 30 cm and linear mass density 0.1 kg/m. which is taut at a tension of 40 N. A small rider (piece of paper) is placed on string at point P as shown. An external vibrating tuning fork is brought near this string and oscillations of rider are carefully observed 1. At which of the following frequencies of turning fork, rider will not vibrate at all :

(A) 100/3 Hz

(B) 50 Hz

(C) 200 Hz

(D) None of these

2. At which of the following frequencies the point P on string will have maximum oscillation amplitude among all points on string :

(A) 200/3 Hz

(B) 100 Hz

(C) 200 Hz

(D) None of these

3. Now if the tension in the string is made 160 N, at which of the following frequencies of turning fork, rider will not vibrate at all

(A) 100/3 Hz

(B) 50 Hz

(C) 200 Hz

(D) None of these

Click to See Solution :
Ans: 1:(C) , 2:(D) , 3:(C)
Sol:

## A sounding body emitting a frequency of 150 Hz is dropped from a height. During its fall under gravity ….

Q: A sounding body emitting a frequency of 150 Hz is dropped from a height. During its fall under gravity it crosses a balloon moving upwards with a constant velocity of 2m/s one second after it started to fall. The difference in the frequency observed by the man in balloon just before and just after crossing the body will be:
(given that -velocity of sound = 300 m/s; g = 10 m/s2)

(a) 12

(b) 6

(c) 8

(d) 4

Ans: (a)

Solution: $\displaystyle f = f_0 ( \frac{v \pm v_0}{v \pm v_s} )$

When approaching :

$\displaystyle f_1 = 150 ( \frac{300 + 2}{300 – 10} )$

When receding :

$\displaystyle f_2 = 150 ( \frac{300 – 2}{300 + 10} )$

$\displaystyle f_1 – f_2 = 12$

## A stationary tuning fork is in resonance with an air column in a pipe . If the tuning fork is moved with a speed of 2 m/s ….

Q: A stationary tuning fork is in resonance with an air column in a pipe . If the tuning fork is moved with a speed of 2 m/s in front of the open end of the pipe and parallel to it , the length of the pipe should be changed for the resonance to occur with moving tuning fork . If the speed of sound in air is 320 m/s , the smallest value of the percentage change required in the length of the pipe is ————–

Ans: (0.62)
Sol: $f \propto \frac{1}{l_1}$

$f = \frac{k}{l_1}$ ……(i)

where l1 = initial length of pipe

$\displaystyle (\frac{v }{v-v_T})f = \frac{k}{l_2}$ …(ii)

Where vT =speed of tuning fork , l2 = new length of pipe

Dividing (i) by (ii)

$\displaystyle \frac{v-v_T}{v} = \frac{l_2}{l_1}$

$\displaystyle 1-\frac{v_T}{v} = \frac{l_2}{l_1}$

$\displaystyle 1- \frac{l_2}{l_1} = \frac{v_T}{v}$

$\displaystyle \frac{l_1 – l_2}{l_1} \times 100 = \frac{v_T}{v} \times 100$

$\displaystyle \frac{l_1 – l_2}{l_1} \times 100 = \frac{2}{320} \times 100 = 0.625$

## The driver of a bus approaching a big wall notices that the frequency of his bus’s horn changes from 420 Hz to 490 Hz ….

Q: The driver of a bus approaching a big wall notices that the frequency of his bus’s horn changes from 420 Hz to 490 Hz when he hears it after it gets reflected from the wall . Find the speed of bus if speed of the sound is 330 m/s

(a) 91 km/h

(b) 61 km/h

(c) 81 km/h

(d) 71 km/h

Click to See Solution :
Ans: (a)
Sol: Frequency appeared at wall

$\displaystyle f’ = \frac{v}{v-v_b}\times f$

$\displaystyle f’ = \frac{330}{330-v_b}\times f$ …(i)

Now Frequency of reflected sound ,

$\displaystyle f” = \frac{330 + v_b}{330}\times f’$

$\displaystyle f” = \frac{330 + v_b}{330}\times \frac{330}{330-v_b}\times f$ from(i)

$\displaystyle f” = \frac{330 + v_b}{330-v_b}\times f$

$\displaystyle 490 = \frac{330 + v_b}{330-v_b}\times 420$

vb = 25.4 m/s = 91 km/h