Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

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

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

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
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring