Question

Two light strings, each of length $l$, are fixed at points A and B on a fixed horizontal rod xy. A small bob is tied by both strings and in equilibrium, the strings are making angle $45^\circ$ with the rod. If the bob is slightly displaced normal to the plane of the strings and released then period of the resulting small oscillation will be:

• A. $2\pi \sqrt{\frac{2\sqrt{2}l}{g}}$
• B. $2\pi \sqrt{\frac{\sqrt{2}l}{g}}$
• C. $2\pi \sqrt{\frac{l}{g}}$
• D. $2\pi \sqrt{\frac{l}{\sqrt{2}g}}$

Answer 1

Answer: (D)

2\pi \sqrt{\frac{l}{\sqrt{2}g}}

Answer 2

The problem describes a bob tied by two light strings of length $l$ each, fixed at points A and B on a horizontal rod. In equilibrium, the strings make an angle of $45^\circ$ with the rod. The bob is then slightly displaced normal to the plane of the strings and released. We need to find the period of the resulting small oscillation.

1. Equilibrium Analysis: In equilibrium, the bob is at a depth $h = l \sin 45^\circ = l/\sqrt{2}$ below the rod. The horizontal distance from the center of the rod to each string's attachment point is $d = l \cos 45^\circ = l/\sqrt{2}$. The vertical components of tension balance the weight: $2T \cos 45^\circ = mg$, so $T = \frac{mg\sqrt{2}}{2}$.

2. Displacement and Restoring Force: When the bob is displaced by a small distance $z$ normal to the plane of the strings (along the z-axis), its new position is $(0, y, z)$. Since the strings are inextensible, the distance from each fixed point (e.g., A at $(-d, 0, 0)$) to the bob must be $l$.

   So, $(-d-0)^2 + (0-y)^2 + (0-z)^2 = l^2$ $d^2 + y^2 + z^2 = l^2$ $(l/\sqrt{2})^2 + y^2 + z^2 = l^2$ $l^2/2 + y^2 + z^2 = l^2 \implies y^2 + z^2 = l^2/2$.

   For small $z$, the vertical position $y = -\sqrt{l^2/2 - z^2} \approx -(l/\sqrt{2})(1 - z^2/l^2)$.

   The forces acting on the bob are its weight $mg$ downwards and the tension $T$ in each string. Due to symmetry, the tension in both strings is equal.

   The restoring force in the z-direction is $F_z = -2T \sin\theta_z$, where $\theta_z$ is the angle the string makes with the y-z plane. More precisely, it's the component of tension in the z-direction.

   The z-component of force from each string is $T \frac{-z}{l}$.

   So, the net restoring force is $F_z = -2T \frac{z}{l}$.

   In the y-direction, for small oscillations, the bob is nearly in equilibrium, so $2T \frac{-y}{l} - mg = 0 \implies T = \frac{mg l}{2|y|}$.

   Substitute $|y| = \sqrt{l^2/2 - z^2}$: $T = \frac{mg l}{2\sqrt{l^2/2 - z^2}}$.

   Now substitute $T$ into the expression for $F_z$: $F_z = -2 \left( \frac{mg l}{2\sqrt{l^2/2 - z^2}} \right) \frac{z}{l} = -\frac{mg z}{\sqrt{l^2/2 - z^2}}$.

   For small oscillations, $z \ll l/\sqrt{2}$, so $\sqrt{l^2/2 - z^2} \approx \sqrt{l^2/2} = l/\sqrt{2}$.

   Thus, $F_z \approx -\frac{mg z}{l/\sqrt{2}} = -\frac{mg\sqrt{2}}{l} z$.

3. Period of Oscillation: The equation of motion is $m \frac{d^2z}{dt^2} = -\left(\frac{mg\sqrt{2}}{l}\right)z$.

   This is a simple harmonic motion equation $m \frac{d^2z}{dt^2} = -kz$, where $k = \frac{mg\sqrt{2}}{l}$.

   The angular frequency is $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{mg\sqrt{2}/l}{m}} = \sqrt{\frac{g\sqrt{2}}{l}}$.

   The period of oscillation is $P = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{l}{g\sqrt{2}}}$.

   This can be rationalized as $P = 2\pi \sqrt{\frac{l\sqrt{2}}{g\sqrt{2}\sqrt{2}}} = 2\pi \sqrt{\frac{\sqrt{2}l}{2g}}$.

   Comparing with the given options, $2\pi \sqrt{\frac{l}{\sqrt{2}g}}$ matches option (4).