CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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?


Your Answer
A
x=2vmk
Your Answer
B
x=1vmk
Correct Answer
C
x=vmk
Your Answer
D
x=3mvk

Solution

The correct option is D x=vmk
In 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=vmk
When mass is at instantaneous rest, position of particle from mean position is A=vmk
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=vmk

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image