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. 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.

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,

\[T = 2\pi\sqrt{\frac{M}{k}}\] 

\[\text { On  substituting  the  values  of  M  and  k  in  the  bove  equation,   we  have: 

\[  T = 2\pi\sqrt{\frac{4}{100}} = \frac{2\pi}{5}  s\] 

\[\text { Frequency  of  the  motion  is  given  by, }\] \[  \frac{1}{T} = \frac{5}{2\pi}  Hz\]

Let v be the velocity of the 1 kg block, at mean position.

\[\text { As  kinetic  energy  is  equal  to  the  potential  energy,   we  can  write: }\] \[\frac{1}{2}m v^2  = \frac{1}{2}k x^2\]

where x = amplitude = 0.1 m

\[\text { Substituting  the  value  of  x  in  above  equation  and  solving  for  v,   we  get: }\] 

\[\left( \frac{1}{2} \right) \times \left( 1 \times v^2 \right) = \left( \frac{1}{2} \right) \times 100 \left( 0 . 1 \right)^2 \] 

\[            v = 1   {ms}^{- 1}                                                          .  .  . \left( 1 \right)\]

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

\[\Rightarrow V = \frac{1}{4}  m/s                                \left[ \text { As } v = 1  {ms}^{- 1} ,\text {  from  equation } (1) \right]\]

Thus, at the mean position, two blocks have a velocity of \[\frac{1}{4} {ms}^{- 1}\]

\[\text { Mean  value  of  kinetic  energy  is  given  as, }\] 

\[KE  \text { at  mean  position }= \frac{1}{2}M V^2 \] 

\[                             = \left( \frac{1}{2} \right) \times 4 \times  \left( \frac{1}{4} \right)^2  = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\] 

At the extreme position, the spring-mass system has only potential energy.

\[PE = \frac{1}{2}k \delta^2  = \frac{1}{2} \times \frac{1}{4}\]

where δ is the new amplitude.

\[\therefore \frac{1}{4} = 100   \delta^2 \] 

\[             = \delta = \sqrt{\left( \frac{1}{400} \right)}\] 

\[           = 0 . 05 \text{ m }  =   5 \text{ cm }\]