Discussion
A mass attached to the bottom end of a vertical massless spring stretches the spring by \(\Delta x\). The system executes oscillation with a time period \(T = 0.2\ s\). The value of \(\Delta x\), in cm, rounded off to the nearest integer is:
[Assume the acceleration due to gravity \(g = 9.8\ m/s^{2}\)]
Answer: D
\(\Delta x\) is about 1 cm.
At equilibrium the spring force balances the weight, \(k\,\Delta x = mg\), so \(m/k = \Delta x/g\). The period of the vertical oscillation is then
\[ \begin{aligned} T &= 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{\Delta x}{g}} \\ \Delta x &= \frac{gT^{2}}{4\pi^{2}} = \frac{9.8 \times 0.04}{39.48} \\ &= 0.0099\ \text{m} \approx 1\ \text{cm} \end{aligned} \]
The mass need not be known: the static stretch alone sets the period, exactly as the length sets the period of a pendulum. The options 2, 3 and 4 cm would give periods of about 0.28, 0.35 and 0.40 s.
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