What is the velocity of a block in SHM involving a spring after a given time?

[Mdhiggenz]

Homework Statement

A 1.6-kg block on a horizontal frictionless surface is attached to a spring whose force constant is 190 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The velocity of the block at time t = 0.40 s is closest to:

Homework Equations

The Attempt at a Solution

ω=√k/m=190/1.6=118.75 rad/s

v=-Aωcos(ωt)

=-0.080*(10.90)cos(10.90*.40)=.30m/s

However the answer is .8m/s

Am I missing a step?

Thank you

 

[Doc Al]

You are using x = Asin(ωt). But that assumes that at t = 0 the mass is at x = 0.

(Pick a form that starts at maximum displacement.)

 

[Mdhiggenz]

I don't quite understand what you are saying,

isnt the maximum displacement .080 which is the amplitude, I am not sure how to incorporate it.

 

[Doc Al]

v=-Aωcos(ωt)

Where did this come from?

When they say "at time t = 0.40 s" they are measuring the time from the moment it was released, which is the point of maximum displacement.

 

[Mdhiggenz]

Which means I use -waSin(wt)

since the graph of sin starts at zero?

 

[Doc Al]

Which means I use -waSin(wt)

since the graph of sin starts at zero?

Yes.