Question

the acceleration of particle in shm at 5cm from mean position is 20 cm/sec2
$sec ^{2}$
then value of angular frequency is
​

Answer 1

Answer:

Explanation:

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 ofx=4cmx=4cmwith 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=2π

WhereTT is the time period we're looking for.

Therefore, the time period is:

T=2πω=π  

2

sT=2πω=π  

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=2πlg−−√T=2πlg

Hence, Option A is correct.

Answer 2

$คภรฬєг⤵$

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 ofx=4cmx=4cmwith 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=2π

WhereTT 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

Hence, correct answer is 2

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