Oscillations - Result Question 58
63. A body is executing S.H.M. When the displacements from the mean position are $4 cm$ and $5 cm$, the corresponding velocities of the body are $10 cm$ per sec and $8 cm$ per sec. Then the time period of the body is
[1991]
(a) $2 \pi sec$
(b) $\pi / 2 sec$
(c) $\pi sec$
(d) $(3 \pi / 2) sec$
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Solution:
- (c) For S.H.M., Velocity,
$v=\omega \sqrt{a^{2}-x^{2}}$ at displacement $x$.
$\Rightarrow 10=\omega \sqrt{a^{2}-16}$
and $8=\omega \sqrt{a^{2}-25}$
Dividing, $\frac{5^{2}}{4^{2}}=\frac{a^{2}-16}{a^{2}-25}=\frac{25}{16}$
or, $16 a^{2}-256=25 a^{2}-625 \Rightarrow 9 a^{2}=369$ $\therefore a^{2}=\frac{369}{9}$
Putting this value in equation (2) mentioned above,
$10=\omega \sqrt{\frac{369}{9}-16} \Rightarrow 10=\omega \sqrt{\frac{225}{9}}$
or, $\omega=\frac{10 \times 3}{15}=2 radian / sec$.
Time period $=\frac{2 \pi}{\omega}=\frac{2 \pi}{2}=\pi sec$.
