29 - Waves and Oscillations Questions Answers

Asked By: SAURABH SHARMA
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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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Ans . [root(3m/m+M)]w0
Asked By: RISHI
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Joshi sir comment
Sir pls solve above 3 questions pls sir
Asked By: KANDUKURI ASHISH
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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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Asked By: TUSHAR GUPTA
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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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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 = 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

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

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