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:

  1. (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 $=\frac{2 \pi}{\omega}=\frac{2 \pi}{2}=\pi sec$.