WBJEE · Physics · Oscillations
A small mass \(m\), attached to one end of a spring with a negligible mass and an unstretched length \(L\), executes virtual oscillation with angular frequency \(\omega_{0}\). When the mass is rotated with an angular speed \(\omega\) by holding the other end of the spring at a fixed point, the mass moves uniformly in a circular path in a horizontal plane. Then the increase in length of the spring during the rotation is
- A \(\frac{\omega^{2} L}{\omega_{0}^{2}-\omega^{2}}\)
- B \(\frac{\omega_{0}^{2} L}{\omega^{2}-\omega_{0}^{2}}\)
- C \(\frac{\omega^{2} L}{\omega_{0}^{2}}\)
- D \(\frac{\omega_{0}^{2} L}{\omega^{2}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\omega^{2} L}{\omega_{0}^{2}-\omega^{2}}\)
Step-by-step Solution
Detailed explanation
The given situation can be shown as From figure, \(K\left(x \sin \theta=m \omega^{2}(L+x) \sin \theta\right.\) \(\Rightarrow \quad K x=\operatorname{m\omega}^{2}(L+x)\)...(i) Also, \(\quad \sqrt{\frac{K}{m}}=\omega_{0}\) \(\Rightarrow \quad K=m \omega_{0}^{2}\) Substituting the…
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