Question

The maximum speed and acceleration of partic!
executing SHM be 20 cm/s and 20 cm/s
respectively, then its time period is

Answer 1

Answer:

Explanation:

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)=−ω2x

Where ω 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=4cm with a negative acceleration of a=−64cms−2.

Therefore:

a(4)=−ω2×4=−64

This implies:

ω=4rads/s

Now,

ω=2πT

Where T is the time period we're looking for.

Therefore, the time period is:

T=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−−√

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

Answer 2

Answer:T= 2π

Explanation:V is maximum when particle is at mean position.

Let V=Awcos(wt) (when x=Asin(wt)

as phase constant is zero particle is at mean position at t=0. So Vmax=Aw=20

|a|=Aw^2sin(wt).

acceleration (a) is max at extreme. So a=Aw*wsin(wT/4)

20=20wsin(2π\T*T/4)

1=wsin(π/2)

1=2π/T

T=2π