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MHT CET · Physics · Laws of Motion

When a mass ' \(\mathrm{m}\) ' is suspended from a spring of length ' \(\ell^{\prime}\), the length of the spring
becomes ' \(\mathrm{L}^{\prime}\). The mass is pulled down by a distance ' \(\mathrm{d}\) ' and released. If the equation of motion of the mass is \(\frac{d^{2} x}{d t^{2}}+\mathrm{P}^{2} x=0\), then \(\mathrm{P}\) is equal to
\((\mathrm{g}=\) acceleration due to gravity \()\)

  1. A \(\frac{\mathrm{L}-\ell}{\mathrm{g}}\)
  2. B \(\frac{\mathrm{g}}{\mathrm{L}-\ell}\)
  3. C \(\sqrt{\frac{\mathrm{g}}{\mathrm{L}-\ell}}\)
  4. D \(\sqrt{\frac{\mathrm{L}-\ell}{\mathrm{g}}}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\sqrt{\frac{\mathrm{g}}{\mathrm{L}-\ell}}\)

Step-by-step Solution

Detailed explanation

INTIAL LENGTH \(=l\)
FINAL LENGTH=L.
NET DISTANCE/LENGTH
PULLED \(=L-l\)
\(\therefore T I M E P E R I O D=2 \pi \sqrt{\frac{L-\ell}{g}}\) allo, \(P=\frac{2 \pi}{T .}\)
\(\begin{aligned} \therefore P=& \frac{g}{L-l} . \\ &(\text { option }-3) . \end{aligned}\)