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Class 11 NEET Physics Oscillations Questions

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Question : 75 of 78

Marks: +1, -0

A body is executing simple harmonic motion. When the displacements from the mean position is

4 \text{ cm}

5 \text{ cm},

the corresponding velocities of the body is

10 \text{ cm} / \text{s}

8 \text{ cm} / \text{s} .

Then the timeperiod of the body is

$2 \pi \text{ s}$
$\frac{\pi}{2} / \text{s}$
$\pi$ sec
$\frac{3 \pi}{2} \text{ s}$

Solution:

For simple harmonic motion velocity

v=\omega \sqrt{a^{2}-x^{2}}

at displacement

x

.

10=\omega \sqrt{a^{2}-16}

.....(i)

8=\omega \sqrt{a^{2}-25}

.....(ii)

\frac{100}{\omega^{2}}=a^{2}-16

... (iii)

\frac{64}{\omega^{2}}=a^{2}-25

.....(iv)

\therefore

Equation

( \text{iii})-( \text{iv})

gives

\frac{36}{\omega^{2}}=9

\Rightarrow\ \omega=2 \text{ rad} / \text{s}

or

T=\frac{2 \pi}{\omega}=\frac{2 \pi}{2}=\pi \text{ s}

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