Question 360: Oscillations

Question

The $x$-$t$ graph of a particle undergoing simple harmonic motion is shown in the figure. The acceleration of the particle at $t=4/3$ s is
![](https://app-content.cdn.examgoal.net/fly/@width/image/1lfrloxnh/d6ed570d-1ecf-43f7-9b07-53e370ee3c03/db4d80d0-cd0a-11ed-8124-cf50c8498624/file-1lfrloxni.png?format=png)

Accepted Answer

The equation of single harmonic motion is $x = A\sin \left( {{{2\pi } \over T}t} ight)$, where A is amplitude and T is the time period of the motion.
From the given figure, $A = 1$ and $T = 8$. Thus,
$x = \sin \left( {{\pi  \over 4}t} ight)$
The acceleration is
$a = {{{d^2}x} \over {d{t^2}}} = {{ - {\pi ^2}} \over {16}}\sin \left( {{\pi  \over 4}t} ight)$
At time $t = {4 \over 3}s$, the acceleration is
$a = {{ - {\pi ^2}} \over {16}}\sin \left( {{\pi  \over 4} 	imes {4 \over 3}} ight) =  - {{\sqrt 3 } \over {32}}{\pi ^2}$ cm/s$^2$

Answer

${{\sqrt 3 } \over {32}}{\pi ^2}$ cm/s$^2$

Answer

${{ - {\pi ^2}} \over {32}}$ cm/s$^2$

Answer

${{  {\pi ^2}} \over {32}}$ cm/s$^2$

Answer

$ - {{\sqrt 3 } \over {32}}{\pi ^2}$ cm/s$^2$

Oscillations

The x-t graph of a particle undergoing simple harmonic motion is shown in the figure. The acceleration of the particle at t=4/3 s is

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