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Oscillations Part 1

Section: Physics

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Question : 19 of 100

Marks: +1, -0

A block of mass

1\,\mathrm{kg}

attached to a spring is made to oscillate with an initial amplitude of

12\,\mathrm{cm}

2\,\mathrm{min}

, the amplitude decreases to

6\,\mathrm{cm}

. Determine the value of the damping constant for this motion.

2=0.693

[17 Mar 2021 Shift 2]

$0.69 \times 10^{2}\,\mathrm{kg/s}$
$3.3 \times 10^{2}\,\mathrm{kg/s}$
$1.16 \times 10^{-2}\,\mathrm{kg/s}$
$5.7 \times 10^{-3}\,\mathrm{kg/s}$

Solution:

Given, mass block,

m=1\,\mathrm{kg}

Initial amplitude,

A_0=12\,\mathrm{cm}

Final amplitude,

A=6\,\mathrm{cm}

The time taken to reduce the amplitude,

t=2\,\mathrm{min}=120\,\mathrm{s}

Using the expression of damped oscillation,

A=A_0 e^{-\;\frac{b}{2m} t}

Substituting the values in the above equation, we get

6=12 e^{-\;\frac{b(120)}{2(1)}}

\Rightarrow \;\; e^{60b}=2

or

60 b=10\log 2

\Rightarrow \;\; b=\;\frac{0.693}{60}=1.16 \times 10^{-2}\,\mathrm{kg/s}

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