Question

The acceleration of a particle in SHM at 5 cm from its mean position is 20 $cm/se{c}^2$. The value of angular frequency in radians/sec will be :

• A. 2
• B. 4
• C. 10
• D. 14

Answer 1

The correct option is A 2

Let's use the known relationship between acceleration, a, and displacement, x, of a simple harmonic mass.

$a\left(x\right)=-\omega 2xa\left(x\right)=-\omega 2x$

Where $\omega \omega$ is the angular frequency of the oscillation and the negative sign shows that the mass is always accelerated towards the centre of oscillation (for a pendulum, this is caused by gravity).

Now we just substitute in your values and see if we can find the angular frequency.

I'm assuming, since you omitted a minus sign in your acceleration or displacement that the mass is at a positive displacement of$x=4cmx=4cm$with a negative acceleration of $a=-64cms-2.a=-64cms-2.$

Therefore:

$a\left(4\right)=-\omega 2\times 4=-64a\left(4\right)=-\omega 2\times 4=-64$

This implies:

$\omega =4rads/s\omega =4rads/s$

Now,

$\omega =2\pi T\omega =2\pi$

Where$TT$ is the time period we're looking for.

Therefore, the time period is:

$T=2\pi \omega ={\pi }^{2}sT=2\pi \omega ={\pi }^{2}s$

This is approximately a period of 1.57 seconds.

As a side note, the time period of oscillation is independent of starting displacement. It is related to the restoring force of the oscillation and/or the mass oscillating.

For instance, for a simple pendulum, the time period is given by:

$T=2\pi lg--\surd T=2\pi lg$

Hence, Option $A$ is correct.