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

It is given that:
Amplitude of simple harmonic motion, x = 0.1 m
Total mass of the system, M = 3 + 1 = 4 kg (when both the blocks move together)
Spring constant, k = 100 N/m
​Time period of SHM $\left(T\right)$ is given by,
$$\begin{gathered} T=2\mathrm{\pi }\sqrt{\frac{M}{k}} \\ \mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{of}M\mathrm{and}k\mathrm{in}\mathrm{the}\mathrm{bove}\mathrm{equation},\mathrm{we}\mathrm{have}: \\ T=2\mathrm{\pi }\sqrt{\frac{4}{100}}=\frac{2\mathrm{\pi }}{5}\mathrm{s} \\ \mathrm{Frequency}\mathrm{of}\mathrm{the}\mathrm{motion}\mathrm{is}\mathrm{given}\mathrm{by}, \\ \frac{1}{T}=\frac{5}{2\mathrm{\pi }}\mathrm{Hz} \end{gathered}$$
Let v be the velocity of the 1 kg block, at mean position.
$$\begin{gathered} \mathrm{As}\mathrm{kinetic}\mathrm{energy}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{potential}\mathrm{energy},\mathrm{we}\mathrm{can}\mathrm{write}: \\ \frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2 \end{gathered}$$
where x = amplitude = 0.1 m
$$\begin{gathered} \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\mathrm{above}\mathrm{equation}\mathrm{and}\mathrm{solving}\mathrm{for}v,\mathrm{we}\mathrm{get}: \\ \left(\frac{1}{2}\right)\times \left(1\times v^2\right)=\left(\frac{1}{2}\right)\times 100{\left(0.1\right)}^2 \\ v=1{\mathrm{ms}}^{-1}...\left(1\right) \end{gathered}$$
When the 3 kg block is gently placed on the 1 kg block, the 4 kg mass and the spring become one system. As a spring-mass system experiences external force, momentum should be conserved.
Let V be the velocity of 4 kg block.
Now,
Initial momentum = Final momentum
∴ 1 × v = 4 × V
$\Rightarrow V=\frac{1}{4}\mathrm{m}/\mathrm{s}\left[\mathrm{As}v=1{\mathrm{ms}}^{-1},\mathrm{from}\mathrm{equation}\left(1\right)\right]$
Thus, at the mean position, two blocks have a velocity of $\frac{1}{4}{\mathrm{ms}}^{-1}$.
$$\begin{gathered} \mathrm{Mean}\mathrm{value}\mathrm{of}\mathrm{kinetic}\mathrm{energy}\mathrm{is}\mathrm{given}\mathrm{as}, \\ \mathrm{KE}\mathrm{at}\mathrm{mean}\mathrm{position}=\frac{1}{2}M{V}^2 \\ =\left(\frac{1}{2}\right)\times 4\times {\left(\frac{1}{4}\right)}^2=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8} \end{gathered}$$
At the extreme position, the spring-mass system has only potential energy.
$\mathrm{PE}=\frac{1}{2}k{\mathrm{\delta }}^2=\frac{1}{2}\times \frac{1}{4}$
where δ is the new amplitude.
$$\begin{gathered} \therefore \frac{1}{4}=100{\delta }^2 \\ =\delta =\sqrt{\left(\frac{1}{400}\right)} \\ =0.05\mathrm{m}=5\mathrm{cm} \end{gathered}$$