Question

There is a system of infinite pulleys and springs. Spring constants are in GP as k, 2k, 4k, 8k... shown in the figure. All the pulleys are massless and frictionless and strings are massless and inextensible. Then time period of oscillation will be? Mass of block is m.

• A. $2\pi\sqrt{\frac{8m}{k}}$
• B. $\pi\sqrt{\frac{8m}{k}}$
• C. $2\pi\sqrt{\frac{4m}{k}}$
• D. $2\pi\sqrt{\frac{m}{8k}}$

Answer 1

Answer: (B)

B

Answer 2

Let $x$ be the downward displacement of the block from its equilibrium position. Let $y_n$ be the downward displacement of the center of the $n$-th movable pulley from its equilibrium position. From the geometry of the pulley system, if the block moves down by $x$, the first movable pulley moves down by $y_1 = x/2$. The center of the first pulley is attached to a string that goes over the second movable pulley. If the first pulley moves down by $y_1$, the second movable pulley moves down by $y_2 = y_1/2 = (x/2)/2 = x/4$. Similarly, $y_3 = y_2/2 = (x/4)/2 = x/8$, and in general, $y_n = x/2^n$.

The springs are attached to the movable pulleys. The $n$-th spring has a spring constant $k_n = 2^{n-1}k$. When the block is displaced by $x$, the $n$-th spring is stretched or compressed by $y_n = x/2^n$. The potential energy stored in the $n$-th spring is $U_n = \frac{1}{2} k_n y_n^2 = \frac{1}{2} (2^{n-1}k) (x/2^n)^2 = \frac{1}{2} 2^{n-1}k \frac{x^2}{2^{2n}} = \frac{1}{2} k \frac{2^{n-1}}{2^{2n}} x^2 = \frac{1}{2} k \frac{1}{2^{n+1}} x^2$.

The total potential energy stored in all the springs is the sum of the potential energies in each spring: $U = \sum_{n=1}^{\infty} U_n = \sum_{n=1}^{\infty} \frac{1}{2} k \frac{1}{2^{n+1}} x^2 = \frac{1}{2} k x^2 \sum_{n=1}^{\infty} \frac{1}{2^{n+1}}$. The sum is a geometric series with the first term (for $n=1$) $a = 1/2^{1+1} = 1/4$ and the common ratio $r = 1/2$. The sum of an infinite geometric series is $S = \frac{a}{1-r} = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$. So, the total potential energy is $U = \frac{1}{2} k x^2 \left(\frac{1}{2}\right) = \frac{1}{4} k x^2$.

The restoring force on the block is $F = -\frac{dU}{dx} = -\frac{d}{dx} \left(\frac{1}{4} k x^2\right) = -\frac{1}{4} k (2x) = -\frac{1}{2} k x$. The equation of motion for the block is $m \frac{d^2x}{dt^2} = F = -\frac{1}{2} k x$. $m \frac{d^2x}{dt^2} + \frac{1}{2} k x = 0$. This is the equation of simple harmonic motion $m \frac{d^2x}{dt^2} + K_{eq} x = 0$, where $K_{eq}$ is the equivalent spring constant. Comparing the equations, we have $K_{eq} = \frac{1}{2} k$.

The angular frequency of oscillation is $\omega = \sqrt{\frac{K_{eq}}{m}} = \sqrt{\frac{k/2}{m}} = \sqrt{\frac{k}{2m}}$. The time period of oscillation is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k/2}} = 2\pi \sqrt{\frac{2m}{k}}$.

Comparing with the options, option (B) is $\pi\sqrt{\frac{8m}{k}} = \pi\sqrt{4 \cdot \frac{2m}{k}} = 2\pi\sqrt{\frac{2m}{k}}$. This matches our result.

Therefore, the correct answer is (B).