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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.

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Solution



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 T is given by,
T=2Ï€MkOn substituting the values of M and k in the bove equation, we have: T=2Ï€4100=2Ï€5 sFrequency of the motion is given by, 1T=52Ï€ Hz
Let v be the velocity of the 1 kg block, at mean position.
As kinetic energy is equal to the potential energy, we can write:12mv2=12kx2
where x = amplitude = 0.1 m
Substituting the value of x in above equation and solving for v, we get:12×1×v2=12×1000.12 v=1 ms-1 ...1
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
⇒V=14 m/s As v=1 ms-1, from equation (1)
Thus, at the mean position, two blocks have a velocity of 14ms-1.
Mean value of kinetic energy is given as,KE at mean position=12MV2 =12×4×142=12×14=18
At the extreme position, the spring-mass system has only potential energy.
PE=12kδ2=12×14
where δ is the new amplitude.
∴14=100 δ2 =δ=1400 =0.05 m = 5 cm

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