Question

A particle is executing simple harmonic motion with a time period of 3 s. At a position where the displacement of the particle is 60% of its amplitude, the ratio of the kinetic and potential energies of the particle is:

• A. 5 : 3
• B. 16 : 9
• C. 4 : 3
• D. 25 : 9

Hint:

N/A

Answer 1

The Correct Option is B

The problem involves calculating the ratio of kinetic energy (KE) to potential energy (PE) for a particle in simple harmonic motion (SHM) at a specific displacement. The displacement given is 60% of the amplitude.

In SHM, the total mechanical energy \(E\) of the system is the sum of kinetic energy \( (KE) \) and potential energy \( (PE) \), both of which vary with displacement. The total energy is given by the equation:

\(E = KE + PE = \frac{1}{2}m\omega^2a^2\)

where:

• \(m\) is the mass of the particle,
• \(\omega\) is the angular frequency,
• \(a\) is the amplitude of vibration.

At a displacement \(x\), which is 60% of the amplitude, we have \(x = 0.6a\).

For a particle in SHM:

\(KE = \frac{1}{2}m\omega^2(a^2 - x^2)\)

\(PE = \frac{1}{2}m\omega^2x^2\)

Using \(x = 0.6a\), substitute into the equations:

\(KE = \frac{1}{2}m\omega^2(a^2 - (0.6a)^2)\)

\(= \frac{1}{2}m\omega^2(a^2 - 0.36a^2)\)

\(= \frac{1}{2}m\omega^2(0.64a^2)\)

\(PE = \frac{1}{2}m\omega^2(0.6a)^2\)

\(= \frac{1}{2}m\omega^2(0.36a^2)\)

Now, find the ratio:

\(\frac{KE}{PE} = \frac{0.64a^2}{0.36a^2} = \frac{0.64}{0.36} = \frac{64}{36} = \frac{16}{9}\)

Thus, the ratio of kinetic energy to potential energy is \(16 : 9\).

Answer 2

We are given a particle executing simple harmonic motion (SHM) with a time period \( T = 3 \, \text{s} \). The displacement of the particle is 60% of its amplitude. We are asked to find the ratio of kinetic and potential energies at this position. 
Step 1: Relationship between displacement, velocity, and energy For SHM, the displacement of the particle is given by: \[ x = A \sin(\omega t), \] where \( A \) is the amplitude, and \( \omega \) is the angular frequency. The angular frequency \( \omega \) is related to the time period by: \[ \omega = \frac{2\pi}{T}. \] The total mechanical energy in SHM is given by: \[ E_{\text{total}} = \frac{1}{2} m \omega^2 A^2. \] This energy is constant, and it is the sum of the kinetic energy (\( K \)) and potential energy (\( U \)): \[ E_{\text{total}} = K + U. \] 
Step 2: Kinetic and potential energies At any point during the motion, the kinetic energy is: \[ K = \frac{1}{2} m \omega^2 (A^2 - x^2), \] and the potential energy is: \[ U = \frac{1}{2} m \omega^2 x^2. \] Now, we are given that the displacement \( x = 0.6A \), so we can substitute this into the equations for \( K \) and \( U \). The kinetic energy at \( x = 0.6A \) is: \[ K = \frac{1}{2} m \omega^2 \left( A^2 - (0.6A)^2 \right) = \frac{1}{2} m \omega^2 \left( A^2 - 0.36A^2 \right) = \frac{1}{2} m \omega^2 \times 0.64A^2. \] The potential energy at \( x = 0.6A \) is: \[ U = \frac{1}{2} m \omega^2 (0.6A)^2 = \frac{1}{2} m \omega^2 \times 0.36A^2. \] 
Step 3: Ratio of kinetic to potential energy Now, we can find the ratio of \( K \) to \( U \): \[ \frac{K}{U} = \frac{\frac{1}{2} m \omega^2 \times 0.64A^2}{\frac{1}{2} m \omega^2 \times 0.36A^2} = \frac{0.64}{0.36} = \frac{16}{9}. \] 
Thus, the ratio of kinetic energy to potential energy is \( 16 : 9 \). Therefore, the correct answer is option (2).