JEE Mains · Physics · STD 11 - 13. oscillations
An object of mass \(m\) is suspended at the end of a massless wire of length \(L\) and area of cross\(-\)section, \(A\). Young modulus of the material of the wire is \(Y\). If the mass is pulled down slightly its frequency of oscillation along the vertical direction is
- A \(f =\frac{1}{2 \pi} \sqrt{\frac{ YA }{ mL }}\)
- B \(f=\frac{1}{2 \pi} \sqrt{\frac{Y L}{m A}}\)
- C \(f=\frac{1}{2 \pi} \sqrt{\frac{m A}{Y L}}\)
- D \(f =\frac{1}{2 \pi} \sqrt{\frac{ mL }{ YA }}\)
Answer & Solution
Correct Answer
(A) \(f =\frac{1}{2 \pi} \sqrt{\frac{ YA }{ mL }}\)
Step-by-step Solution
Detailed explanation
An elastic wire can be treated as a spring with \(K =\frac{ Y A }{\ell}\) \(T =2 \pi \sqrt{\frac{ m }{ k }}\) \(f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{Y A}{m \ell}}\)
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