Question

A particle executing SHM along a straight line has zero velocity at points $A$ and $B$ whose distance from $O$ on the same line $OAB$ are $a$ and $b$, respectively. If the velocity at the mid point between $A$ and $B$ is $v$, then its time period is

• A. $\frac{\pi (b+a)}{v}$
• B. $\pi \left(\frac{b-a}{v}\right)$
• C. $\left(\frac{b+a}{2v}\right)$
• D. $\left(\frac{b-a}{2v}\right)$

Answer 1

Answer: (B)

According to the question,

$\therefore$ Amplitude
$=\frac{\text{ Distance travelled by the particles }}{2}(A$ to $B)$
Amplitude of particles executing
simple harmonic motion (SHM) along a straight
line $AB$ is $(a)=\frac{b-a}{2}$.
Velocity of particle, $v=$ Amplitude $\times$ Oscillation frequency
$\therefore v=a\omega$
$v=\left(\frac{b-a}{2}\right)\omega \text{ or }\omega =\frac{2v}{b-a}$ ....(i)
$\therefore$ Time period, $T=\frac{2\pi }{\omega }$
Putting the value of $\omega$ from Eq. (i) to above formula,
$\Rightarrow T=\frac{2\pi }{2v}\times (b-a)\Rightarrow T=\frac{b-a}{v}\pi$
So, the time period of a particle executing SHM
along a straight line from points $A$ to $B$ is,
$T=\frac{b-a}{v}\pi$