# 22 - Waves and Oscillations Questions Answers

while measuring the speed of sound by performing a resonance column expt , a student gets d first resonance conditn at a column length of 18 cm during winter . Repeating d same expt during summer , she measure d column length to be x cm for the second resonance. THEN

1)18>X

2)X>54

3) 54>X>36

4) 36>X>18

ANS (2)

AIEEE 2008

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

since l α λ α v/n n is frequency

on increasing temperature, v increases so λ increases

for first resonance l = λ/4

and for third resonance L = 3λ^{'}/4

and λ^{' }> λ so x will be more than 54

when a tuning fork vibrates with 1.0 m or 1.05m long wire of a sonometer, 5 beats per second are produced in each case. what will be the frequency of the tuning fork ??

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

for sonometer wire

l = λ/2

or 2l = v/n here n represents frequency

or n = v/2l

for first wire of 1 m length n= v/2

for second wire of 1.05 m length n=v/2.1

let frequency of tuning fork = n_{1}

so according to the given condition v/2 = n_{1}+ 5

and v/2.1 = n_{1} - 5

solve now

TWO PENDULUM HAVE TIME PERIODS T &5T/4. THEY STARTS SHM AT THE SAME TIME FROM THE MEAN POSITION . WHAT WILL BE THE PHASE DIFFERENCE B/W THEM AFTER D BIGGER PENDULUM COMPLETED ONE OSCILLATn ??

ans 90^{0}

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

When bigger pendulum will complete its one oscillation, the smaller one will complete [1+(1/4)] oscillation. so phase difference = 2π*(T/4)/T = π/2

a cylinderical tube open at both ends has a fundamental frequency f in air the tube is dipped vertically in H_{2}O so that half of it is in H2O the fundamental frequency of the air column is

1)f/2

2)3f/4

3)2f

4) f

aieee 2012

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

for first case l = λ/2 so l = v/2f

for second case l/2 = λ/4 so l/2 = v/4f

so f in both case are same.

QUES TWO IDENTICAL SOUNDS s_{1 }&s_{2} reach at a point P in phase. resultant loudness at a point P is ndB higher than loudness of s_{1} the value of n =?

ANS) 6

**Asked By: SARIKA**

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**Joshi sir comment**

Let intensity of one source = I

then intensity at a point where two waves meet in phase = I+I+2√I√Icos0 = 4I

now loudness = 10logI/I_{0} due to a single source

and loudness at the point where 2 waves meet in phase = 10log4I/I_{0 }= 10logI/I_{0}+ 10log4

since 10log4 = 6 so increment = 6

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*a uniformrope of lenth 10 m and mass 12 kg hangs verticaly from a rigid support . a block of mass 4 kg is attached to the free end of the rope . atrans verse wave of wavelenth 00.03 m is prodused at the lower end . wavelenth of pulse when it reaches upper end will be?*

**Asked By: SHUBHAM VED**

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**Joshi sir comment**

at top point tension in the string = 16g

and at bottom point T = 4g

and v = √(T/m) here m is mass per unit length which is same throughout the rope.

first calculate v at both the points

frequency will be same throughout

so use the formula v = nλ twice

once at bottom and once at top point

then by diving these 2 formulae we will get the result

try it.

2 comparable masses m_{1 }and m_{2 }r orbiting about their bodycenter o at distance r_{1} and r_{2} from their body center respectively with a common period T. the squar of time period ?

**Asked By: SHUBHAM VED**

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**Joshi sir comment**

According to the diagram

Gm_{1}m_{2}/(r_{1}+r_{2})^{2} = m_{1}r_{1}ω^{2} = m_{2}r_{2}ω^{2}

compare first part with either second or third part of the given equation for getting ω^{2} then find T^{2}

how can we calculate ratio of intensities of two sound waves having unequal amplitudes (one having double the amplitude of the other - both are sinusodial waves) and the graph is displacement vs time .

**Asked By: AMIT DAS**

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**Joshi sir comment**

Intensity of a wave depends on two power of amplitude and two power of frequency. Check the graph carefully for getting idea about frequency and details of amplitude are given in the question.

the ratio of intensities between two coherent sound sources is 4:1. the difference of loudness in decibel between maximum and minimum intensities, when they interfere in space is

(1) 10 log2

(2) 20 log3

(3) 10 log3

(4) 20 log2

**Asked By: HIMANSHU MITTAL**

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**Joshi sir comment**

I_{1}/I_{2 }= 4/1 so a_{1}/a_{2} = 2/1 so a_{max}/ a_{min} = 2+1/2-1 = 3/1 so I_{max}/ I_{min}= 9/1

now loudness L = 10 log(I/I_{0})

so difference of loudness = L_{1}-L_{2} = 10 log (9I/I) = 10 log9 = 20 log3

two coherent light sources A and B with seperation 2λ are placed the x-axis symmetrically about the origin. they emit light of wavelength λ. obtain the position of maximas on a circle of large radius R lying in the xy plane and with the center at origin

**Asked By: HIMANSHU MITTAL**

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**Joshi sir comment**

for maxima path diffrence should be integral multiple of λ.

It is given that two coherent light sources A and B with seperation 2λ are placed on the x-axis symmetrically about the origin, and we have to get the points of maxima in the circle so at the point where circle cuts the X axis (path diffrence = 2λ) and where circle cuts the Y axis (path difference = 0) will be the point of maxima. Besides it one point will also be present in the circle in first quadrant where path difference will be λ so it will be a maxima also. it means total 8 points are there