# 25 - Waves and Oscillations Questions Answers

A small ring is threaded on an inextensible frictionless cord of length 2l. The ends of the cord are fixed to a horizontal ceiling. In equilibrium, the ring is at a depth h below the ceiling. Now the ring is pulled aside by a small distance in the vertical plane containing the cord and released. If period of small oscillations of the ring is kπ, find k. Given,

.

**Asked By: ANOUSHKA**

**submit your answer**

**Joshi sir comment**

small ring is threaded on an inextensible frictionless cord of length 2l.

The ends of the cord are fixed to a horizontal ceiling. In equilibrium, the

ring is at a depth h below the ceiling. Now the ring is pulled aside by a

small distance in the vertical plane containing the cord and released.

Find period of small oscillations of the ring. Acceleration of free fall is g.

**Asked By: ABHISHEK KHANDELWAL**

**submit your answer**

**Joshi sir comment**

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**

**submit your answer**

**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**

**submit your answer**

**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**

**submit your answer**

**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**

**submit your answer**

**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**

**submit your answer**

**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

##
*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**

**submit your answer**

**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**

**submit your answer**

**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}