# 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.

Joshi sir comment

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,

.

Joshi sir comment
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.

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

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 ??

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 = n1

so according to the given condition v/2 = n1+ 5

and v/2.1 = n1 - 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 900