Question

A block of mass M = 3m is connected to a spring of mass m and oscillates in simple harmonic motion on a horizontal, frictionless track (figure). The force constant of the spring is k, and the equilibrium length is $l$. Assume all portions of the spring oscillate in phase and the velocity of a segment dx is proportional to the distance x from the fixed end; that is, $v_x = (x/l)v$. Also, notice that the mass of a segment of the spring is dm = (m/l)dx.

• A. The kinetic energy of the system when the block has a speed v is $\frac{5}{3}mv^2$
• B. The kinetic energy of the system when the block has a speed v is $\frac{4}{3}mv^2$
• C. The period of oscillation is $2\pi\sqrt{\frac{10m}{3k}}$
• D. The period of oscillation is $2\pi\sqrt{\frac{8m}{3k}}$

Answer 1

Answer: (A), (C)

A, C

Answer 2

The problem asks us to find the kinetic energy of the system and the period of oscillation for a block-spring system where the spring itself has mass.

1. Kinetic Energy of the Block: The mass of the block is $M = 3m$, and its speed is $v$. The kinetic energy of the block ($KE_{block}$) is given by: $KE_{block} = \frac{1}{2} M v^2 = \frac{1}{2} (3m) v^2 = \frac{3}{2} m v^2$

2. Kinetic Energy of the Spring: The spring has a total mass $m$ and equilibrium length $l$. Consider a small segment of the spring of length $dx$ at a distance $x$ from the fixed end. The mass of this segment ($dm$) is given by: $dm = \frac{m}{l} dx$ The velocity of this segment ($v_x$) is given as proportional to the distance $x$ from the fixed end: $v_x = \left(\frac{x}{l}\right)v$ The kinetic energy of this small segment ($dKE_{spring}$) is: $dKE_{spring} = \frac{1}{2} dm v_x^2 = \frac{1}{2} \left(\frac{m}{l} dx\right) \left(\frac{x}{l}v\right)^2$ $dKE_{spring} = \frac{1}{2} \frac{m}{l} \frac{x^2}{l^2} v^2 dx = \frac{1}{2} \frac{m v^2}{l^3} x^2 dx$ To find the total kinetic energy of the spring ($KE_{spring}$), we integrate $dKE_{spring}$ from $x=0$ to $x=l$: $KE_{spring} = \int_{0}^{l} \frac{1}{2} \frac{m v^2}{l^3} x^2 dx = \frac{1}{2} \frac{m v^2}{l^3} \int_{0}^{l} x^2 dx$ $KE_{spring} = \frac{1}{2} \frac{m v^2}{l^3} \left[\frac{x^3}{3}\right]_{0}^{l} = \frac{1}{2} \frac{m v^2}{l^3} \left(\frac{l^3}{3} - 0\right)$ $KE_{spring} = \frac{1}{2} \frac{m v^2}{3} = \frac{1}{6} m v^2$

3. Total Kinetic Energy of the System: The total kinetic energy ($KE_{total}$) is the sum of the kinetic energy of the block and the kinetic energy of the spring: $KE_{total} = KE_{block} + KE_{spring} = \frac{3}{2} m v^2 + \frac{1}{6} m v^2$ To add these fractions, find a common denominator (6): $KE_{total} = \frac{9}{6} m v^2 + \frac{1}{6} m v^2 = \frac{10}{6} m v^2 = \frac{5}{3} m v^2$ This matches option (A).

4. Effective Mass of the System: For simple harmonic motion, the total kinetic energy can be expressed as $\frac{1}{2} m_{eff} v^2$, where $m_{eff}$ is the effective mass of the oscillating system. Comparing this with our calculated total kinetic energy: $\frac{1}{2} m_{eff} v^2 = \frac{5}{3} m v^2$ $m_{eff} = \frac{2 \times 5}{3} m = \frac{10}{3} m$ This effective mass is consistent with the standard result for a spring with mass $m_s$ attached to a block of mass $M_b$, which is $m_{eff} = M_b + \frac{m_s}{3}$. Here, $M_b = 3m$ and $m_s = m$, so $m_{eff} = 3m + \frac{m}{3} = \frac{10m}{3}$.

5. Period of Oscillation: The period of oscillation ($T$) for a spring-mass system is given by the formula: $T = 2\pi\sqrt{\frac{m_{eff}}{k}}$ Substitute the effective mass $m_{eff} = \frac{10}{3} m$: $T = 2\pi\sqrt{\frac{\frac{10}{3}m}{k}} = 2\pi\sqrt{\frac{10m}{3k}}$ This matches option (C).

Both options (A) and (C) are correct.