Simple harmonic motion finding velocity trouble

  • Thread starter Eats Dirt
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  • #1
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Homework Statement



A block with mass m =6.6 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.26 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.4 m/s. The block oscillates on the spring without friction.


at time 0.39 s what is the speed of the block

Homework Equations



v= -Awsin(wt)

The Attempt at a Solution



i sub A= 0.716681695
w= 6.139406135
t= 0.39 into the equation and get 2.990262698

my calculator is on rads and i really dont understand what is going wrong in this calculation i found the other variables using the given values
 

Answers and Replies

  • #2
998
15
It looks to me that at t = 0s, the velocity is 4.4m/s.

But the equation that you are using i.e. v= -Awsin(wt)
is inconsistent with this because it gives v = 0 at t = 0,
 
  • #3
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It looks to me that at t = 0s, the velocity is 4.4m/s.

But the equation that you are using i.e. v= -Awsin(wt)
is inconsistent with this because it gives v = 0 at t = 0,

which equation did you use?
 
  • #4
92
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which equation did you use?

ok so you use cos instead but how do you know which one to use? the sin equation or the cos equation?
 
  • #5
998
15
An equation consistent with the conditions of the problem is v = Awcos(wt)
 
  • #6
998
15
It all depends at which point the time is taken as 0.
The displacement equation is x = Asinwt if the stopwatch is started at the equilibrium i.e.t = 0 at x = 0.

But the displacement is x = Acoswt if the stopwatch is started at the positive extreme end of the oscillation i.e. x = A when t = 0.

Then the corresponding equation for the velocity is used.
 
  • #7
92
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An equation consistent with the conditions of the problem is v = Awcos(wt)

yes but how would i know which one to use sin or cos? is there a general rule for telling which one should be used? or telling when there should be a phase shift etc?

thankyou for the help so far!
 
  • #8
998
15
Sometimes the information given in the problem gives at which point in the shm the time t = 0. For example in the problem quoted, the mass is given a speed from the equilibrium position. And then one is asked to find something later on after that. So one can take t = 0 when the particle was at equilibrium i.e when displacement x = 0.

Otherwise one must make an assumption about at which point t = 0 and must remain consistent with that assumption.
 
  • #9
92
0
Sometimes the information given in the problem gives at which point in the shm the time t = 0. For example in the problem quoted, the mass is given a speed from the equilibrium position. And then one is asked to find something later on after that. So one can take t = 0 when the particle was at equilibrium i.e when displacement x = 0.

Otherwise one must make an assumption about at which point t = 0 and must remain consistent with that assumption.

ok thankyou so much! although i am confused a little bit because when the box is pushed on the timer starts, is the block not at the equilibrium when it is pushed on in this problem?
 
  • #10
998
15
It WAS at equilibrium BEFORE t = 0 but then AT t = 0 it was given a velocity.
 

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