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A particle starts SHM from the mean position. Its amplitude is A and total energy E. At an instant its kinetic energy is 3E/4. Its displacement at this instant is

A. A/√2
B. A2
C. √3A/2
D. Zero

User Souljacker
by
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1 Answer

2 votes

Final answer:

In Simple Harmonic Motion (SHM), the displacement of a particle at the instant when its kinetic energy is 3E/4 is A/√2. Option A is correct.

Step-by-step explanation:

A particle undergoing Simple Harmonic Motion (SHM) starts from the mean position with an amplitude of A and total energy E. At an instant when its kinetic energy is 3E/4, the displacement of the particle is A/√2.

In SHM, the total energy remains constant and is the sum of the kinetic energy and potential energy. At any given instant, the kinetic energy is given by K = 1/2mv^2, where m is the mass and v is the velocity. So, if the kinetic energy is 3E/4, we can say that K = 3/4 * E.

Since the total energy E remains constant, the potential energy at this instant can be calculated by subtracting the kinetic energy from the total energy: U = E - K = E - (3E/4) = E/4.

According to the energy equation of SHM, the potential energy can be expressed as U = 1/2kx^2, where k is the spring constant and x is the displacement from the mean position. Solving for x, we get x = √(2U/k) = √(2(E/4)/k) = √(E/2k).

Therefore, the displacement of the particle at the instant when its kinetic energy is 3E/4 is A/√2.

User Kimchy
by
7.8k points
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