1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A block of mass m, attached to a spring of spring constant k, oscillates on a smooth horizontal table. The other end of the spring is fixed to a wall. The block has a speed v, when the spring is at its natural length. How far will the block move on the table before coming to an instantaneous rest?

A
x=2vmk
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
x=1vmk
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
x=vmk
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
x=3mvk
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is C x=v√mkIn SHM, speed of the block is maximum at mean position and speed is zero at extreme position. For given case, particle is at mean position, when spring is in its natural length. Maximum velocity in SHM v=Aω Angular frequency of the oscillation is ⇒ω=√km ⇒A=vω ⇒A=v√mk When mass is at instantaneous rest, position of particle from mean position is A=v√mk Hence option 'c' is correct OR From conservation of mechanical energy, Ki+Ui=Kf+Uf Initial kinetic energy is Ki=12mv2 As spring is in its natural length, initial potential energy Ui=0 For mass to be at rest, final kinetic energy is Kf=0 As spring stretches by x, final potential energy Uf=12kx2 kx2=mv2 x=v√mk

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program