Question

If two simple pendulums first of bob mass \( M_1 \) and length \( l_1 \), and \( M_2 \) and \( l_2 \), Given \( M_1 = M_2 \) and \( l_1 = 2l_2 \), If the vibrational energies of both are same, then which of the following is correct?

• A. Amplitude of \( B \) is smaller than \( A \)
• B. Amplitude of \( B \) is greater than \( A \)
• C. Amplitude will be same
• D. None of the above

Hint:

For pendulums with the same energy, the amplitude is inversely related to the length of the pendulum. A shorter pendulum has a smaller amplitude for the same energy.

Answer 1

The Correct Option is A

The vibrational energy of a simple pendulum is given by: \[ E = \frac{1}{2} m g h \] where \( h \) is the height change corresponding to the amplitude of oscillation.
Since the vibrational energies are the same for both pendulums, the height change and thus the amplitude for the pendulum with the shorter length \( l_2 \) (pendulum \( B \)) will be smaller than the amplitude for the pendulum with the longer length \( l_1 \) (pendulum \( A \)).
Thus, the correct answer is (a).