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MHT CET · Physics · Oscillations

A small mass 'm' is suspended at the end of a wire having (negligible mass) length 'L' and cross-sectional area 'A'. The frequency of oscillation for the S.H.M. along the vertical line is \((\mathrm{Y}=\) Young's modulus of the wire)

  1. A \(\frac{1}{2 \pi}\left(\frac{\mathrm{YA}}{\mathrm{mL}}\right)^{\frac{1}{2}}\)
  2. B \(\frac{2 \pi \mathrm{YA}}{\mathrm{mL}}\)
  3. C \(\frac{\mathrm{YA}}{2 \pi \mathrm{mI}}\)
  4. D \(2 \pi\left(\frac{\mathrm{YA}}{\mathrm{mL}}\right)^{\frac{1}{2}}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{2 \pi}\left(\frac{\mathrm{YA}}{\mathrm{mL}}\right)^{\frac{1}{2}}\)

Step-by-step Solution

Detailed explanation

If \(x\) is the extension of the wire then
\(P=\left(\frac{Y A}{L}\right) x \quad F=k x \quad \text { where } k=\frac{Y A}{L}\)
frequency of oscillation is given by
\(f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{Y A}{m L}}\)