Question

Two spherical bob of masses \(M_A\) and \(M_B\) are hung vertically from two strings of length \(l_A\) and \(l_B\) respectively. They are executing SHM with frequency relation  \(f_A=2f_B\), Then:

• A. \(l_A=\frac {l_B}{4}\)
• B. \(l_A=4l_B\)
• C. \(l_A=2l_B\) & \(M_A=2M_B\)
• D. \(l_A=\frac {l_B}{2}\) & \(M_A=\frac {M_B}{2}\)

Answer 1

The Correct Option is A

To determine the relationship between the string lengths \(l_A\) and \(l_B\) given the frequency relationship \(f_A = 2f_B\) for two spherical bobs executing Simple Harmonic Motion (SHM), we will use the formula for the frequency of a simple pendulum: f = \frac{1}{2\pi} \sqrt{\frac{g}{l}}, where \(g\) is the acceleration due to gravity and \(l\) is the length of the string.

1. Given: \(f_A = 2f_B\). This can be equated using the frequency formula: \frac{1}{2\pi} \sqrt{\frac{g}{l_A}} = 2 \cdot \frac{1}{2\pi} \sqrt{\frac{g}{l_B}}.
2. Canceling out common terms and simplifying, we arrive at: \sqrt{\frac{g}{l_A}} = 2 \sqrt{\frac{g}{l_B}}.
3. Squaring both sides gives: \frac{g}{l_A} = 4 \cdot \frac{g}{l_B}.
4. This simplifies to: \frac{1}{l_A} = \frac{4}{l_B}, which implies l_A = \frac{l_B}{4}.

Thus, the correct relation between the lengths of the strings is \(l_A = \frac{l_B}{4}\).