Question

The acceleration of particle in shm at 5 cm from its mean position is 20 cm per second square the value of angular frequency is

Answer 1

Well this is a more simple problem than it may first appear to be.

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

a(x)=−ω2xa(x)=−ω2x

Where ωω 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(4)=−ω2×4=−64a(4)=−ω2×4=−64

This implies:

ω=4rads/sω=4rads/s

Now,

ω=2πTω=2πT

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

Therefore, the time period is:

T=2πω=π2sT=2πω=π2s

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=2πlg−−√T=2πlg

Where ll is the length of the pendulum and gg is the gravitational field strength (10 N/kg or 10 ms−2ms−2 on Earth's surface).

Answer 2

Hi there..

I hope it's the right answer..