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