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A mass of 2kg is attached to the spring ...

A mass of 2kg is attached to the spring of spring constant `50Nm^(-1)`. The block is pulled to a distance of 5 cm from its equilibrium position at `x=0` on a horizontal frictionless surface from rest at t=0. Write the expression for its displacement at anytime t.

A

x=0.05sin5t m

B

x=0.05cos5t m

C

x=0.5sin5t m

D

x=5sin5t m

Text Solution

Verified by Experts

The correct Answer is:
A
Here, `m=2kg,k=50N" "m^(-1),A=5cm=0.05m`
the block executes SHM. Its angular frequency is given by
`omega=sqrt((k)/(m))=sqrt((50" N "m^-1)/(2kg))=5" rad "s^(-1)`
Since the time noted from the equilibrium position, its displacement at any timee t is given by
`x=Asinomegat=0.05sin5t` m
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