Question

In a ballistic galvanometer, \( \theta_0 \) be the throw in the absence of damping and \( \theta_1 \) be the first throw after \( \frac{T}{4} \) sec (where \( T \) is the time period of oscillations). The correct relation between \( \theta_0 \), \( \theta_1 \) & logarithmic decrement \( \lambda \) of the galvanometer will be:

• A. \( \theta_1 = \theta_0 (1 + \lambda/2) \)
• B. \( \theta_0 = \theta_1 \left( 1 + \frac{\lambda}{2} \right) \)
• C. \( \theta_1 = \theta_0 e^{\lambda/2} \)
• D. \( \theta_1 = \theta_0 e^{-\lambda/2} \)

Hint:

Damping causes an exponential decay in oscillations: \[ \theta_n = \theta_0 e^{-n\lambda} \]

Answer 1

The Correct Option is D

The damping in a ballistic galvanometer follows an exponential decay:
\[ \theta_n = \theta_0 e^{-n\lambda} \] For the first throw \( \theta_1 \) after \( \frac{T}{4} \):
\[ \theta_1 = \theta_0 e^{-\lambda/2} \]