JEE Main 2020 · 6 September · morning · Single correct · +4 / −1
An object of mass m is suspended at the end of a massless wire of length L and area of crosssection 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 = 1/2π√(YA/mL)
- (b) f = 1/2π√(mL/YA)
- (c) f = 1/2π√(YL/mA)
- (d) f = 1/2π√(mA/YL)
Answer and solution
Answer (a) f = 1/2π√(YA/mL)
Solution
Physics sir · 3 steps
The stretched wire is just a spring: find its spring constant from Young modulus, then use the usual formula.
- Wire as spring. If the wire stretches by Δ L, the definition of Young modulus gives Y = (F/A)/(Δ L/L). So the restoring force is F = YA/L Δ L, which is proportional to the stretch, exactly like a spring.
k = YA/L
- Gravity drops out. Gravity only fixes the equilibrium position, where the wire is already stretched by mg/k. Pull the mass a further x and the extra restoring force is kx, so the motion is SHM with this same k.
- So. f = 1/2π√(k/m) = 1/2π√(YA/mL), which is option (a).
The idea to keep. Any elastic rod or wire behaves like a spring with k = YA/L. A thicker or stiffer wire is stiffer, and a longer wire is softer. Then use f = 1/2π√(k/m).
Answer from the NTA final answer key. Solution by PrepWiser.