30 - 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.
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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,
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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.
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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
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 ??
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