Q: A bob of mass M is hung using a string of length l. A mass m moving with a velocity u pierces through the bob and emerges out with velocity u/3 horizontally . The frequency of small oscillations of the bob, considering A as amplitude is,
(a) $\displaystyle \frac{1}{2\pi} \sqrt{\frac{3 m u}{2 M A}} $
(b) $\displaystyle \frac{1}{2\pi} \sqrt{\frac{2 m }{3 M A}} $
(c) $ \displaystyle \frac{1}{2\pi} (\frac{2 m u}{3 M A}) $
(d) $ \displaystyle \frac{1}{2\pi} (\frac{3 m u}{2 M A}) $
Click to See Answer :
Ans: (c)
Sol: mu + 0 = MV + m(u/3)
MV = 2mu/3
⇒ V= 2mu/3M
Vmax = ωA
⇒ ω = Vmax/A = 2mu/3MA
T = 2π/ω = 2π(3MA/2mu)
⇒ ν = 1/T