Question

The ratio of the average value of kinetic energy to that of potential energy for an SHM is:

• A. 2:1
• B. 1:1
• C. 1:2
• D. 1:4

Hint:

In SHM, the average kinetic energy is equal to the average potential energy over a complete cycle, leading to a ratio of \( 1:1 \).

Answer 1

The Correct Option is B

In Simple Harmonic Motion (SHM), the total mechanical energy is conserved and is equally distributed between the kinetic energy (\( K \)) and the potential energy (\( U \)) at any point in time. The total energy \( E \) of an object performing SHM is given by: \[ E = K + U \] At any position during the SHM: 1. Kinetic Energy (\( K \)) is given by: \[ K = \frac{1}{2} m v^2 \] Where \( m \) is the mass of the object and \( v \) is its velocity at that position. 2. Potential Energy (\( U \)) is given by: \[ U = \frac{1}{2} k x^2 \] Where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. At extreme positions (when \( x = A \), the amplitude), the velocity is zero, so all the energy is stored as potential energy, and at the mean position (when \( x = 0 \)), all the energy is converted to kinetic energy. Now, the average values of kinetic energy and potential energy over a complete cycle of SHM can be derived as: - The average kinetic energy \( \langle K \rangle \) is: \[ \langle K \rangle = \frac{1}{2} E \] - The average potential energy \( \langle U \rangle \) is: \[ \langle U \rangle = \frac{1}{2} E \] Therefore, the ratio of average kinetic energy to average potential energy is: \[ \frac{\langle K \rangle}{\langle U \rangle} = \frac{\frac{1}{2} E}{\frac{1}{2} E} = 1:1 \] Thus, the ratio of the average kinetic energy to the average potential energy for an SHM is \( 1:1 \).