Question

Two point bodies of masses $m$ and $3m$ are connected by a massless spring of spring constant $k = m\omega_0^2$ and kept on a frictionless horizontal surface. The spring is extended by a small distance $l$ over its natural length at time $t = 0$ and then released so that the masses execute simple harmonic motion. The maximum speed of the particle with mass $m$ is given by

• A. $\frac{\sqrt{3}\omega_0 l}{2}$
• B. $\frac{2\omega_0 l}{\sqrt{3}}$
• C. $\frac{\omega_0 l}{\sqrt{3}}$
• D. $\frac{3\omega_0 l}{4}$

Hint:

Using conservation of momentum and conservation of energy at the point of maximum speed (equilibrium position) yields the same result:
\[ \frac{1}{2} k l^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} (3m) v_2^2 \]
With $m v_1 = 3m v_2 \implies v_2 = v_1 / 3$.

Answer 1

The Correct Option is A

Step 1: Picture the two-mass spring.
Masses $m$ and $3m$ are joined by a spring on a frictionless table. With nothing pushing from outside, the centre of mass stays put and the two masses oscillate about it.

Step 2: Use the reduced mass shortcut.
A two-body spring behaves like a single mass on a spring, where that single mass is the reduced mass \[ \mu = \frac{m \cdot 3m}{m + 3m} = \frac{3m}{4}. \]

Step 3: Find the angular frequency.
With $k = m\omega_0^2$, \[ \omega = \sqrt{\frac{k}{\mu}} = \sqrt{\frac{m\omega_0^2}{3m/4}} = \sqrt{\frac{4}{3}}\,\omega_0 = \frac{2\omega_0}{\sqrt{3}}. \]

Step 4: Split the initial stretch between the masses.
The stretch $l$ is shared so that each mass moves inversely to its mass. The light mass $m$ takes the bigger share, \[ A_m = l \cdot \frac{3m}{m + 3m} = \frac{3l}{4}. \]

Step 5: Get the maximum speed of mass $m$.
For simple harmonic motion the top speed is amplitude times angular frequency, \[ v_{\max} = A_m \, \omega = \frac{3l}{4} \cdot \frac{2\omega_0}{\sqrt{3}}. \]

Step 6: Tidy up the expression.
\[ v_{\max} = \frac{6 l \omega_0}{4\sqrt{3}} = \frac{3 l \omega_0}{2\sqrt{3}} = \frac{\sqrt{3}\,\omega_0 l}{2}. \] \[ \boxed{v_{\max} = \dfrac{\sqrt{3}\,\omega_0 l}{2}} \]