MHT CET · Physics · Oscillations
A block of mass ' \(\mathrm{m}\) ' attached to one end of the vertical spring produces extension \(\mathrm{x}^{\prime}\). If the block is pulled and released, the periodic time of oscillation is
- A \(2\pi\sqrt{\frac{2 x}{g}}\)
- B \(2 \pi\sqrt{\frac{\mathrm{x}}{\mathrm{g}}}\)
- C \(2 \pi \sqrt{\frac{\mathrm{x}}{2 \mathrm{~g}}}\)
- D \(2 \pi \sqrt{\frac{\mathrm{x}}{4 \mathrm{~g}}}\)
Answer & Solution
Correct Answer
(B) \(2 \pi\sqrt{\frac{\mathrm{x}}{\mathrm{g}}}\)
Step-by-step Solution
Detailed explanation
(A)
\(\mathrm{mg}=-\mathrm{kx} \quad \therefore \mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}\)
\(\omega^{2}=\frac{k}{m}\)
\(\left(\frac{2 \pi}{T}\right)^{2}=\frac{k}{m}\)
\(\frac{4 \pi^{2}}{T^{2}}=\frac{k}{m}\)
\(\therefore T^{2}=\frac{4 \pi^{2} m}{k}=\frac{4 \pi^{2} m x}{m g} \Rightarrow T=2 \pi \sqrt{\frac{x}{g}}\)
\(\mathrm{mg}=-\mathrm{kx} \quad \therefore \mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}\)
\(\omega^{2}=\frac{k}{m}\)
\(\left(\frac{2 \pi}{T}\right)^{2}=\frac{k}{m}\)
\(\frac{4 \pi^{2}}{T^{2}}=\frac{k}{m}\)
\(\therefore T^{2}=\frac{4 \pi^{2} m}{k}=\frac{4 \pi^{2} m x}{m g} \Rightarrow T=2 \pi \sqrt{\frac{x}{g}}\)
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