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

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

AP EAMCETAP EAMCET 2019

Solution:

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