Question

The angular velocities of three bodies in simple harmonic motion are \(\omega_{1},\omega_{2},\omega_{3}\) with their respective amplitudes as \(A_{1},A_{2},A_{3}\). If all the three bodies have same mass and velocity, then

• A. \(A_1^2\omega_1^2 = A_2^2\omega_2^2 = A_3^2\omega_3^2\)
• B. \(A_1^2\omega_1 = A_2^2\omega_2 = A_3^2\omega_3\)
• C. \(A_1\omega_1^2 = A_2\omega_2^2 = A_3\omega_3^2\)
• D. \(A_1\omega_1 = A_2\omega_2 = A_3\omega_3\)

Hint:

In SHM, maximum velocity \(v_{\text{max}} = A\omega\). When velocities are equal, amplitudes and angular frequencies are inversely related.

Answer 1

The Correct Option is D

To solve this problem, we need to apply the concepts related to simple harmonic motion (SHM) and equate the parameters given the conditions provided. Let's analyze the given information and the properties of SHM step-by-step:

1. The question states that the angular velocities of three bodies in SHM are \(\omega_1\), \(\omega_2\), and \(\omega_3\) respectively, and their corresponding amplitudes are \(A_1\), \(A_2\), and \(A_3\).
2. All three bodies have the same mass and velocity. Since velocity in SHM can be maximum at the mean position, the expression for maximum velocity (\(V_{\text{max}}\)) in SHM is given by: \(V = A\omega\).
3. This implies that when the maximum velocity is constant for the same mass, the product of amplitude and angular velocity also remains constant. Therefore, the relation becomes: \(A_1\omega_1 = A_2\omega_2 = A_3\omega_3\).

Based on the analysis, the correct answer must be \(A_1\omega_1 = A_2\omega_2 = A_3\omega_3\). The remaining options diverge from the derived relationship, making them incorrect. Here, the primary condition was maintaining the same velocity, which directly relates to the product of amplitude and angular frequency for particle motion in SHM.

Thus, the complete understanding of concepts of SHM and careful alignment with given conditions leads us to conclude that the correct option is the fourth one.