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Oscillations Part 1

Section: Physics

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Question : 22 of 100

Marks: +1, -0

Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass

=500 \text{g}

, decay constant

=20 \text{g}/\text{s}

, then how much time is required for the amplitude of the system to drop to half of its initial value?

(\ln 2=0.693)

[16 Mar 2021 Shift 2]

$34.65 \text{s}$
$17.32 \text{s}$
$0.034 \text{s}$
$15.01 \text{s}$

Solution:

Given,

Mass,

m=500 \text{g}

Decay constant,

\lambda=20 \text{g}/\text{s}

For damped oscillation,

A=A_{0} e^{-\frac{b bt}{2m}}

....(i)

Here,

b=

damped constant

=

decay constant

As per question, the amplitude of the system is dropped to half of its initial value.

So, Eq. (i) becomes

\;\frac{A_0}{2}=A_{0} e^{-\frac{b t}{2m}} \Rightarrow \;\frac{1}{2}= e^{-\frac{b t}{2m}}

\Rightarrow \;\; 2= e^{\frac{b t}{2m}}

\Rightarrow \;\; 2= e^{\frac{b t}{2m}}

Taking log on both sides of above equation, we get

\log 2=\log\left(e^{\frac{b t}{2m}}\right)

\Rightarrow \;\; \log 2= \frac{b t}{2m} \Rightarrow t= \frac{2m}{b} \log 2

=\;\frac{2 \times 500 \times 0.693}{20}

[ \because \log 2=0.693]

\Rightarrow t=34.65 \text{s}

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