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Question

For the damped oscillator shown, the mass m of the block is 200 g, k = 90 N māˆ’1 and the damping constant b is 40 g sāˆ’1. What is the period of oscillation, time taken for its amplitude of vibrations to drop to half of its initial value.


A
1s, 6.93s
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B
0.3s, 6.93s
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C
0.6s, 6.93s
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D
0.3s, 3s
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Solution

The correct option is B 0.3s, 6.93s

Here km =90×0.2=18 kg2 s2

km=4.283 kg s1

Given value of b=0.04 kg s1

We can see that b<<km

So we can say the angular frequency of oscillation

ωkm

T2πmk=2×π×0.290=0.3s

The amplitude depends on time as

A(t)=A0ebt2m

So when amplitude drops to half it's initial value we have

A(t)=A02 A02=A0ebt2m

12=ebt2m

ln2=bt2m

t=2m ln2b

=2×0.2×ln20.04

=6.93s


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