Question

A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N m−1. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.
Figure

Answer 1

Answer:

Here is your answer......

Answer 2

Explanation:

To find the frequency:

We can write $f=\left(\frac{1}{2 \pi}\right) \sqrt{\frac{k}{m_{(\text {combined mass})}}}=\frac{2.5}{\pi}=0.8 \mathrm{Hz}$ where $k=100$ is the spring constant.

For block of mass = 1kg, $f_{0}=\frac{5}{\pi} \mathrm{Hz}$.

To find the amplitude of the motion:

By the law of conservation of linear momentum at mean position, $\omega A=\frac{1}{4} \omega_{0} A_{0}$, where $A_{0}$ is the amplitude and $f_0$ is the frequency of simple harmonic motion of 1kg block. This implies $f A=\frac{1}{4} f_{0} A_{0}$.  Thus, $A=\frac{f_{0} A_{0}}{4 f}=0.05 \mathrm{m}$.  Therefore, the amplitude of motion is 0.05m.