Question

Two blocks connected by an inextensible string and connected to two springs in relaxed state are kept on a smooth floor as shown. The 2 kg block is pushed by a distance of 1 cm against right spring and released.

• A. The maximum compression in left spring is $\frac{1}{\sqrt{2}}$cm.
• B. The time period of oscillation is $\frac{\pi}{5}$sec.
• C. The string is always taut (except at mean position)
• D. When right block first comes to rest, left spring is at natural length.

Answer 1

Answer: (B)

The time period of oscillation is $\frac{\pi}{5}$sec.

Answer 2

The system can be treated as a single effective oscillator with a total mass $M = m_1 + m_2 = 1 \text{ kg} + 2 \text{ kg} = 3 \text{ kg}$ and an effective spring constant $k_{eff} = k_1 + k_2 = 200 \text{ N/m} + 100 \text{ N/m} = 300 \text{ N/m}$. The time period of oscillation for such a system is given by the formula $T = 2\pi \sqrt{\frac{M}{k_{eff}}}$. Substituting the values, we get $T = 2\pi \sqrt{\frac{3 \text{ kg}}{300 \text{ N/m}}} = 2\pi \sqrt{\frac{1}{100}} = 2\pi \times \frac{1}{10} = \frac{\pi}{5}$ seconds.

Let's analyze the other options: (A) If the string is always taut, the system oscillates as a single mass. The displacement from equilibrium is $y(t) = -0.01 \cos(\omega t)$, where $\omega = \sqrt{\frac{k_1+k_2}{m_1+m_2}} = 10$ rad/s. The maximum compression in the left spring occurs when $y_1$ is minimum. If $y_1 = y_2 = y$, then the maximum compression is $|y_{min}| = |-0.01| = 0.01$ m or 1 cm. Option A states $\frac{1}{\sqrt{2}}$cm, which is incorrect.

(C) The tension in the string can be calculated. For the 1 kg block, $m_1 \ddot{y}_1 = -k_1 y_1 + T$. Assuming $y_1=y_2=y$, $T = m_1 \ddot{y} + k_1 y = m_1 (-\omega^2 y) + k_1 y = (k_1 - m_1 \omega^2) y$. With $k_1=200$, $m_1=1$, $\omega=10$, $T = (200 - 1 \times 10^2)y = 100y$. Since $y(t) = -0.01 \cos(10t)$, $T(t) = 100(-0.01 \cos(10t)) = -\cos(10t)$. The tension is negative when $\cos(10t) > 0$, which means the string becomes slack. Thus, the string is not always taut.

(D) The right block first comes to rest when its velocity is zero. $\dot{y}_2 = \frac{d}{dt}(-0.01 \cos(10t)) = 0.1 \sin(10t)$. Setting $\dot{y}_2 = 0$, the first time is $10t = \pi$, so $t = \pi/10$. At this time, $y_2(\pi/10) = -0.01 \cos(\pi) = 0.01$ m. If the string were taut, $y_1 = y_2 = 0.01$ m. The left spring is at its natural length when $y_1=0$. Since $y_1=0.01$ m, the left spring is stretched, not at its natural length.