Question 276: Oscillations

Question

A particle performs linear S.H.M. When the displacement of the particle from mean position is 3 cm and 4 cm , corresponding velocities are $8 \mathrm{~cm} / \mathrm{s}$ and $6 \mathrm{~cm} / \mathrm{s}$ respectively. Its periodic time is

Accepted Answer

We have $v=\omega \sqrt{A^2-x^2}$
Substituting the given values in the above equation we get,
$8=\omega \sqrt{\mathrm{A}^2-9}$ $\ldots$ (i) $\left(\because x=3 \mathrm{~cm}, \mathrm{v}=8 \mathrm{cms}^{-1}ight)$
$6=\omega \sqrt{A^2-16}$ ...(ii) $\left(\because x=4 \mathrm{~cm}, \mathrm{v}=6 \mathrm{~cms}^{-1}ight)$
Diving equation (i) and (ii),
$$\begin{array}{ll} 
& \frac{8}{6}=\frac{\sqrt{\mathrm{A}^2-9}}{\sqrt{\mathrm{~A}^2-16}} \
& \frac{4}{3}=\frac{\sqrt{\mathrm{A}^2-9}}{\sqrt{\mathrm{~A}^2-16}} \
	herefore \quad & \frac{16}{9}=\frac{\mathrm{A}^2-9}{\mathrm{~A}^2-16} \
	herefore \quad & 16 \mathrm{~A}^2-256=9 \mathrm{~A}^2-81 \
& 7 \mathrm{~A}^2=175 \
	herefore \quad & \mathrm{~A}^2=\frac{175}{7}
\end{array}$$
Substituting this value of $A^2$ in equation (ii) and solving we get
$$\begin{aligned}
6 & =\omega \sqrt{\frac{175}{7}-16} \
6 & =\omega \sqrt{\frac{175-112}{7}}=\omega \sqrt{9} \
6 & =3 \omega \
\omega & =2 \
	herefore \quad \omega & =\frac{2 \pi}{T}=2 \
	herefore \quad T & =\pi 	ext { seconds }
\end{aligned}$$

Answer

$2 \pi \mathrm{~s}$

Answer

$\pi \mathrm{s}$

Answer

$3 \pi \mathrm{~s}$

Answer

$4 \pi \mathrm{~s}$

Oscillations

A particle performs linear S.H.M. When the displacement of the particle from mean position is 3 cm and 4 cm , corresponding velocities are 8 \mathrm{~cm} / \mathrm{s} and 6 \mathrm{~cm} / \mathrm{s} respectively. Its periodic time is

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