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Finding postion, velocity and accelration as functions of time (SHM)

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Homework Statement



A 1.00-kg glider attached to a spring with a force constant 16.0 N/m oscillates on a frictionless, horizontal air track. At t = 0, the glider is released from rest at x = -2.50 cm (that is, the spring is compressed by 2.50 cm).

Find the position, velocity, and acceleration as functions of time. (Where position is in m, velocity is in m/s, acceleration is in m/s2, and t is in s. Use the following as necessary: t.)

Homework Equations



x(t)=Acos(ωt+ φ)

v(t)= dx/dt

a(t)=d^2x/dt^2

ω=√(k/m)

The Attempt at a Solution




I know that the answer is x(t)= .025cos(4t+π) (then taking the derivatives respectively). What I can't figure out however is why φ=π. I thought if the particle is at its maximum position (x=A) at t=0, φ=0?
 

Answers and Replies

  • #2
vela
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You're right that if t=0 and φ=0, the oscillator starts at x=A. The amplitude A is a positive number. Perhaps that's the bit of info you're missing. In this problem, you know the oscillator starts at the other extreme, x=-A, where A = 0.025 m.
 
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I copied the question word for word. It was the third question in the problem however. It asked me to find the period and the maximum velocity and acceleration. But I don't see how that would matter in the third question.
 
  • #4
vela
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I have no idea what you're asking.
 
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I'm asking why φ=π in this question when I thought it should be zero because of the given information.
 
  • #6
HallsofIvy
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My first thought, seeing a "glider" and a spring was that the glider was being lauched into the air by the spring! But that clearly is not the case- your "glider" is simply a mass attached to the spring so that it is pushed out and pulled back by the spring. Since the spring has spring constant 16 N/m, the force, at distance x from the neutral point is -16x Newton's. That is, you have [itex]mdx^2/dt^2= -16x[/itex] or [itex]dx^2/dt^2= -16x/m[/itex] which is just [itex]dx^2/dt^2= -16x[/itex] since the mass is 1 kg. The fact that the glider is " released from rest at x = -2.50 cm" means that x(0)= -0.0250 (because the spring constant is given in Newtons per meter, you want x in meters)and dx/dt(0)= 0.

Since you apparently are given "x(t)=Acos(ωt+ φ)", just replace x in each of dx/dt= -16x, x(0)= -2.5, dx/dt(0)= 0. You will get three equations to solve for the three parameters, A, ω, and φ.

(The condition [itex]d^2x/dt^2= -16x[/itex] must be true for all x. Comparing the right and left sides will tell you what ω is.)
 
  • #7
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My first thought, seeing a "glider" and a spring was that the glider was being lauched into the air by the spring! But that clearly is not the case- your "glider" is simply a mass attached to the spring so that it is pushed out and pulled back by the spring. Since the spring has spring constant 16 N/m, the force, at distance x from the neutral point is -16x Newton's. That is, you have [itex]mdx^2/dt^2= -16x[/itex] or [itex]dx^2/dt^2= -16x/m[/itex] which is just [itex]dx^2/dt^2= -16x[/itex] since the mass is 1 kg. The fact that the glider is " released from rest at x = -2.50 cm" means that x(0)= -0.0250 (because the spring constant is given in Newtons per meter, you want x in meters)and dx/dt(0)= 0.

Since you apparently are given "x(t)=Acos(ωt+ φ)", just replace x in each of dx/dt= -16x, x(0)= -2.5, dx/dt(0)= 0. You will get three equations to solve for the three parameters, A, ω, and φ.

(The condition [itex]d^2x/dt^2= -16x[/itex] must be true for all x. Comparing the right and left sides will tell you what ω is.)

At first they asked me to find the period of this glider, so to find ω I just did 2π/T
 
  • #8
vela
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I'm asking why φ=π in this question when I thought it should be zero because of the given information.
I explained that earlier. When t=0, you get x(0) = A cos φ where x(0) = -0.025 m and A = +0.025 m. Solve for φ.
 
  • #9
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I explained that earlier. When t=0, you get x(0) = A cos φ where x(0) = -0.025 m and A = +0.025 m. Solve for φ.
Oh! i see it now. You'll get -1=cosφ making φ=π

Thank you and sorry for overlooking what you told me I overlooked :uhh:
 
  • #10
gneill
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I know that the answer is x(t)= .025cos(4t+π) (then taking the derivatives respectively). What I can't figure out however is why φ=π. I thought if the particle is at its maximum position (x=A) at t=0, φ=0?
Take advantage of the trig identity cos(θ + π) = -cos(θ).

In this problem you want the initial displacement to be -A at time t = 0. By adding π to the argument of the cosine function you accomplish changing the sign.
 

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