# 29 - Waves and Oscillations Questions Answers

The quality factor of a sonometer wire is 2 x 10³. On plucking it makes 240 per second. Calculate the time in which the amplitude decreases to half the initial value.

**Asked By: SINOVIYA V K**

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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**

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**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**

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**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**

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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