Two blocks with masses m1=1kg and m2=2 kg are connected by a spring constant k=24 Nm−1 and placed on a frictionless horizontal surface. The block m1 is imparted an initial velocity v0=12 cms −1 to the right, the amplitude of oscillation is
Question
Two blocks with masses m1=1kg and m2=2 kg are connected by a spring constant k=24 Nm−1 and placed on a frictionless horizontal surface. The block m1 is imparted an initial velocity v0=12 cms −1 to the right, the amplitude of oscillation is
Solution
The problem describes a system of two blocks connected by a spring, where one block is given an initial velocity. This is a problem of simple harmonic motion, and we are asked to find the amplitude of the oscillation.
The total energy of the system is conserved, and is given by the sum of the kinetic and potential energies. Initially, all the energy is kinetic (since the blocks are at rest and the spring is unstretched), and is given by (1/2)m1v0^2.
At the maximum amplitude, all the energy is potential (since the blocks are momentarily at rest and the spring is stretched to its maximum), and is given by (1/2)kA^2, where A is the amplitude.
Setting these two equal gives:
(1/2)m1v0^2 = (1/2)kA^2
Solving for A gives:
A = sqrt((m1*v0^2)/k)
Substituting the given values:
A = sqrt((1 kg * (12 cm/s)^2) / 24 N/m)
Note: We need to convert cm/s to m/s by multiplying by 0.01.
A = sqrt((1 kg * (0.12 m/s)^2) / 24 N/m)
A = sqrt((0.0144 kg*m^2/s^2) / 24 N/m)
A = sqrt(0.0006 m^2)
A = 0.0245 m
So, the amplitude of the oscillation is approximately 0.0245 meters.
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