30 - Waves and Oscillations Questions Answers

in a large room, a person receives direct sound wave from a source 120m away from him. he also receives wave from the same source which reaches him after being reflected from 25m high ceiling at apoint midway between them. the two waves interfere constructively for wavelength of

(1) 20, 20/3, 4

(2) 10, 5, 3.33

(3) 10, 20, 30

(4) 15, 25, 35

Asked By: HIMANSHU MITTAL
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Joshi sir comment

path difference between two waves one coming directly (path length = 120 m) and other coming after reflection (path length = 130), you can get this by geometry 

so path difference = 10 m

for constructive interference path difference should be a multiple of wavelength so (2) is answer

a source of sound and an observer are moving with velocities v = (2i + 4j)m/s and v = (6i - 3j)m/s. if frequency of source is 400 Hz, then frequency observed by observer will be

 

Asked By: HIMANSHU MITTAL
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Joshi sir comment

dot product of the two velocities given = 0, means observer and source are moving perpendicularly, so observer will observe the original frequency.  

two waves are represented by y = Acosωt and y = Asin(ωt - π/6), superimpose on each other. the amplitude of the rsultant wave is

Asked By: HIMANSHU MITTAL
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Joshi sir comment

first convert cos into sin then calculate phase difference, its value will be π/2 + π/6 = 2π/3

now use the formula for resultant amplitude

which of the following phenomenon cannot be observed with sound waves?

(1) diffraction

(2) polarisation

(3) reflection

(4) interference

Asked By: HIMANSHU MITTAL
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Joshi sir comment

polarisation

a source of sound of frequency f emits sound wave with speed v is at rest. if an observer is moving towards source with speed Vo(V not), then wavelength of sound waves observed by the observer will be

(1) v/f

(2) (v - Vo)/f

(3) [v(v - Vo)]/[f(v + Vo)]

(4) [v(v + Vo)]/[f(v - Vo)]

Asked By: HIMANSHU MITTAL
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Joshi sir comment

since velocity and frequency are changing proportionally so wavelength will be constant = v/f

a sonometer wire supports a 4kg load and vibrates in fundamental mode with a tuning fork of frequency 416 Hz. the length of the wire between the bridges is now doubled. in order to maintain fundamental mode the load should be changed to

Asked By: HIMANSHU MITTAL
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Joshi sir comment

for sonometer wire 

in fundamental mode l = λ/2 = v/2f       here l, λ and f are length , wavelength and frequency respectively

now v = √(T/µ)      here T and µ are tension and mass per unit length

so finally we will get l = √(T/µ)/2f

make two equations by using this formula and given conditions

in the two equations f and µ will be same

in a stationery wave, there is no energy current, but there is energy density. (TRUE/FALSE)

Asked By: HIMANSHU MITTAL
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Joshi sir comment

TRUE

at STP, the speed of sound in hydrogen is 1324ms‾¹ then the spped of sound in air (at STP) is

Asked By: HIMANSHU MITTAL
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Joshi sir comment

formula for speed of sound is v = √(γRT/M)

you have to form two equations for the two mediums, for both the mediums γ will be same because Hydrogen and Air both are diatomic gases, T will also be same for both the cases are taken in STP, only M will be different

a person is standing between two vertical walls. he heard the echo of his sound after t = 1s and t = 1.5s. distance between walls is (speed of sound in air = 340ms‾¹

Asked By: HIMANSHU MITTAL
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Joshi sir comment

time taken by the sound to reach upto the first wall = 0.5 sec

similarly time taken by the sound to reach upto the second wall = 0.75

so total time = 1.25 sec. so distance between the walls = 340 * 1.25 = 425

which of the following is minimum audible wavelength at room temperature?

(1) 20mm

(2) 20cm

(3)20m

(4) 20km

Asked By: HIMANSHU MITTAL
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Joshi sir comment

maximum audible frequency = 20000 Hz

Velocity of sound = 340 m/s

so minimum audible wavelength = 340/20000 m = 34000/20000 cm = 340/20 mm = 17 mm approx 20 mm

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