Question

The the velocity versus time graph of a particle executing SHM is shown below the maximum acceleration of particle is

Answer 1

Given:

Velocity versus time graph of a particle executing SHM is shown.

To find:

Max acceleration?

Calculation:

From the graph, we can say :

$\rm \: v = 2 \cos( \omega t)$

Now, integrating:

$\rm \implies \: \dfrac{dx}{dt} = 2 \cos( \omega t)$

$\rm \implies \: dx = 2 \cos( \omega t) \: dt$

$\rm \implies \: \displaystyle\int \rm dx = 2 \int\cos( \omega t) \: dt$

$\rm \implies \: x = \dfrac{2}{ \omega} \sin( \omega t)$

• So, amplitude is $\dfrac{2}{\omega}$.

Now, max acceleration:

$\rm \: a_{max} = { \omega}^{2}A$

$\rm \implies \: a_{max} = { \omega}^{2} \times \dfrac{2}{ \omega}$

$\rm \implies \: a_{max} = 2\omega$

$\rm \implies \: a_{max} = 2 \times \dfrac{2\pi}{T}$

• From graph, time period is 2π.

$\rm \implies \: a_{max} = 2 \times \dfrac{2\pi}{2\pi}$

$\rm \implies \: a_{max} = 2 \: m {s}^{ - 2}$

So, max acceleration is 2 m/s².