Question 363: Oscillations

Question

A particle is performing a linear simple harmonic motion of amplitude ' $A$ '. When it is midway between its mean and extreme position, the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

Accepted Answer

In linear simple harmonic motion, the velocity of particle is given by
$$v=\omega \sqrt{A^2-x^2}\quad	ext{.... (i)}$$
where, $\omega=$ angular frequency
$A=$ maximum displacement of amplitude
and $\quad x=$ displacement from mean position.
The acceleration of a particle in simple harmonic motion, (SHM) is given by
$$\begin{aligned}
& a=\omega^2 x \quad	ext{.... (ii)}\
& 	ext { Here, } \quad x=\frac{A}{2} \
& 	ext { Also, } \quad v=a \quad	ext{(given)}\
& \omega \sqrt{\left(A^2-x^2ight)}=\omega^2 x 	ext { [from Eqs. (i) and (ii), we get] } \
& \Rightarrow \sqrt{\left(A^2-\frac{A^2}{4}ight)}=\omega 	imes \frac{A}{2} \Rightarrow \frac{\sqrt{3} A}{2}=\omega 	imes \frac{A}{2} \
& \Rightarrow \quad \frac{2 \pi}{T}=\sqrt{3} \quad\left[\because \omega=\frac{2 \pi}{T}ight] \
& \Rightarrow \quad T=\frac{2 \pi}{\sqrt{3}} \mathrm{~s}
\end{aligned}$$

Answer

$\frac{2 \pi}{\sqrt{3}} \mathrm{~s}$

Answer

$\frac{\sqrt{3}}{2 \pi} s$

Answer

$2 \pi \sqrt{3} \mathrm{~s}$

Answer

$\frac{1}{2 \pi \sqrt{3}} \mathrm{~s}$

Oscillations

A particle is performing a linear simple harmonic motion of amplitude ' A '. When it is midway between its mean and extreme position, the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

Choose the correct answer: