Question

A 5.0-kg object is suspended from the ceiling by a strong spring, which stretches 0.20 m when the object is attached. The object is lifted 0.060 m from this equilibrium position and released. Part A Determine the amplitude of the resulting simple harmonic motion. Express your answer with the appropriate units. A

Answer 1

Answer:

Time period will be 0.897 sec

Explanation:

We have given mass of object m = 5 kg

Spring stretches by a distance of 0.2 m

So x = 0.2 m

Spring force will be equal to weight of the object

So $mg=kx$

$5\times 9.8=k\times 0.2$

$k=245N/m$

Now angular velocity is given $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{245}{5}}=7rad/sec$

So time period $T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{7}=0.897sec$

Answer 2

Amplitude of the resulting simple harmonic motion will still be 0.897secs.

What is Amplitude?

Amplitude serves as maximum displacement or distance moved by a point on a vibrating body.

We were given mass of object as 5kg

distance of the spring x= 0.2m

we can get the spring constant K using the formula; K=mg/x

Then K= ( 5*9.8)/0.2

=245N/m

The period of the is oscillation can be calculated as w= sqr k/m

= 245/5 = 7 rads/sec.

We can calculate the period as T= (2π/7)= 0.897 secs.

Therefore, amplitude of the resulting simple harmonic motion will still be 0.897secs.

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