Two springs with negligible masses and force constant of k1=200Nm−1 and k2=160Nm−1 are attached to the block of mass m = 10 kg as shown in the figure. Initially the block is at rest, at the equilibrium position in which both springs are neither stretched nor compressed. At time t =0, sharp impulse of 50 Ns is given to the block with a hammer along the spring.Calculate δ if the amplitude of SHM is (5/δ) m.
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is 6.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
T=2πmK1+K2=2π10360=π3sThe maximum velocity is always at equilibrium position since at any other point there will be a restoring force attempting to slow the mass.∴ Vmax= impulse mass =5010=5ms−1A= amplitude =Vmaxω=5(2π)π3=56⇒δ=6