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Question: A \(1\;{\text{kg}}\) block is executing simple harmonic motion of amplitude \(0.1\;{\text{m}}\) on a...

A 1  kg1\;{\text{kg}} block is executing simple harmonic motion of amplitude 0.1  m0.1\;{\text{m}} on a smooth horizontal surface under the restoring force of a spring of spring constant100  Nm−1100\;N{m^{ - 1}}. A block of mass 3  kg3\;{\text{kg}} is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the amplitude of the motion.
A. 0.5  m0.5\;m
B. 0.05  m0.05\;m
C. 5  m5\;m
D. 0.02  m0.02\;m

Explanation

Solution

Hint: Simple Harmonic Motion is the motion of the object possesses a condition in which restoring force is directly proportional to the displacement of the object. The direction of force is in the inverse direction of displacement of the body. Initially, kinetic energy possessed by the body equals the restoring energy generated by the block by using the condition the velocity of the block is calculated. Using the law of conservation of momentum and energy conservation can calculate the amplitude of a block.

Useful formula:
The expression for finding kinetic energy is
K.E=12mv2K.E = \dfrac{1}{2}m{v^2}
Where, mm is the mass of the block and vv is the velocity of the block.

The expression for finding potential energy is
P.E=12kδ2P.E = \dfrac{1}{2}k{\delta ^2}
Where, kk is the spring constant and δ\delta is the deflection of the spring.

Given data:
The mass of the block executing simple harmonic motion is m1=1  kg{m_1} = 1\;{\text{kg}}
The Spring constant is k=100  Nm−1k = 100\;N{m^{ - 1}}
The amplitude of the spring is x=0.1  mx = 0.1\;{\text{m}}
The mass of another block is m2=3  kg{m_2} = 3\;{\text{kg}}

Complete step by step solution:
Then kinetic energy possessed by the block equals restoring energy.
That is 12m1v2=12kx2\dfrac{1}{2}{m_1}{v^2} = \dfrac{1}{2}k{x^2}
Substitute all the values in the above equation.
12(1  kg)v2=12(100  Nmm−1)(0.1)2 v2=1  ms−1 v=1  ms−1  \dfrac{1}{2}(1\;kg){v^2} = \dfrac{1}{2}(100\;Nm{m^{ - 1}}){(0.1)^2} \\\ {v^2} = 1\;m{s^{ - 1}} \\\ v = 1\;m{s^{ - 1}} \\\
The velocity of the block is v=1  ms−1v = 1\;m{s^{ - 1}}
After the 3  kg3\;kg block is gently placed on the 1  kg1\;{\text{kg}} then let, total mass of the system mtotal=4  kg{m_{total}} = 4\;{\text{kg}} block and the spring be one system. For this mass spring system, there is so much external force (when oscillation takes place). The momentum should be conserved. Consider the mass mtotal{m_{total}} has a velocity uu
The expression for momentum conservation is
m1v=mtotalu{m_1}v = {m_{total}}u
Substitute all the values in the above equation.

1  kg×1  ms−1=4  kg×u u=14  ms−1  1\;{\text{kg}} \times 1\;m{s^{ - 1}} = 4\;kg \times u \\\ u = \dfrac{1}{4}\;m{s^{ - 1}} \\\

Now new kinetic energy of the block is
K.E=12mtotalu2K.E = \dfrac{1}{2}{m_{{\text{total}}}}{u^2}
Substitute all the values in the above equation.
K.E=12×4  kg×(14)2 K.E=18  J  K.E = \dfrac{1}{2} \times 4\;{\text{kg}} \times {\left( {\dfrac{1}{4}} \right)^2} \\\ K.E = \dfrac{1}{8}\;J \\\
When the block reaches the mean position energy conversion takes place that is kinetic energy equal to the potential energy.

K.E=P.E 18  J=12kδ2 14  J=100  Nm−1(δ2) δ2=1400  m δ=0.05  m  K.E = P.E \\\ \dfrac{1}{8}\;J = \dfrac{1}{2}k{\delta ^2} \\\ \dfrac{1}{4}\;J = 100\;N{m^{ - 1}}({\delta ^2}) \\\ {\delta ^2} = \sqrt {\dfrac{1}{{400}}} \;{\text{m}} \\\ \delta = 0.05\;{\text{m}} \\\

Thus, the amplitude is δ=0.05  m\delta = 0.05\;{\text{m}}

Note: While the system reaches the mean position, the kinetic energy possessed by the spring converted into potential and amplitude of block depends on the total mass, spring constant and velocity of the block. By using that data, we can find out the frequency and time period of the block.