Discussion
Two masses \(m_{1}\) and \(m_{2}\) are attached to the extreme ends of a massless spring of force constant k as shown in figure below. The system is at rest on a horizontal surface. The vibrational frequency w of the system is :
Answer: A
The frequency is \(\sqrt{\frac{k(m_{1}+m_{2})}{m_{1}m_{2}}}\).
The equations of motion are \(m_{1}\ddot{x}_{1} = k(x_{2} - x_{1})\) and \(m_{2}\ddot{x}_{2} = -k(x_{2} - x_{1})\). The centre of mass moves uniformly, and the extension \(x = x_{2} - x_{1}\) obeys
\[ \begin{aligned} \ddot{x} &= -k\left(\frac{1}{m_{1}} + \frac{1}{m_{2}}\right)x \\ &= -\frac{k}{\mu}x \\ \mu &= \frac{m_{1}m_{2}}{m_{1} + m_{2}} \end{aligned} \]
So the system oscillates like a single reduced mass \(\mu\) on the spring, with \(w = \sqrt{k/\mu}\), which is the keyed expression. Check: if \(m_{2} \to \infty\) (one end fixed) it reduces to \(\sqrt{k/m_{1}}\).
A difference \(m_{1} - m_{2}\) has no physical basis and would give zero or imaginary frequencies for equal masses, and the inverted forms have the dimensions of time, not frequency.
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