Simple Harmonic Oscillator Problem

In summary, the problem is asking for the speed of a mass oscillating on a Slinky, given its position is described by the equation 18.5 cm cos[ 18.0 s-1t] and t = 0.360 s. The formula v(t)=-Aωsin(ωt+θ) can be used to find the velocity, but the result must be converted to speed by taking the magnitude of the velocity. The final answer is 65.1 cm/s.
  • #1
Hibbs
2
0

Homework Statement



The position of a mass that is oscillating on a Slinky (which acts as a simple harmonic oscillator) is given by 18.5 cm cos[ 18.0 s-1t]. What is the speed of the mass when t = 0.360 s?

Homework Equations


x(t)=Acos(ωt+θ)
v(t)=-Aωsin(ωt+θ)


The Attempt at a Solution



I used the formula, v(t)=-Aωsin(ωt+θ) because you basically have everything you need such as:
A=18.5cm
ω=18.0s^-1
t=0.360s

What I get is:
v(0.360)=-(18.5cm)(18.0)sin(18.0*0.360)
to get -65.1cm/s
which isn't the right answer.
Please tell me where I went wrong!
 
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  • #2
Speed is never negative.
 
  • #3
Hibbs said:

Homework Statement



The position of a mass that is oscillating on a Slinky (which acts as a simple harmonic oscillator) is given by 18.5 cm cos[ 18.0 s-1t]. What is the speed of the mass when t = 0.360 s?

The highlighted formula makes little sense to me. First, the argument of the cos must be dimensionless but here it looks like it's time. 18.0 s must be the phase term but what's with the s? Is 18.0 in radians, deg or ? The "1" in front of t must be √(k/m), k = spring const. & m = mass, aka ω. The phase is due to the fact that this mass had initial velocity and displacement. Anyway, no way do I see that ω = 18.0.

Weird! I guess you could go

x = 18.5cos(18 - t) cm
x' = -18.5sin(18 - t) cm/s since I guess ω = 1 rad/s;
so x'(t=0.36) = -18.5sin(18 - 0.36) = -18.5sin(17.64) cm/s.

BTW v can be negative. Speed can't.
 
  • #4
Hibbs said:

Homework Statement



The position of a mass that is oscillating on a Slinky (which acts as a simple harmonic oscillator) is given by 18.5 cm cos[ 18.0 s-1t]. What is the speed of the mass when t = 0.360 s?

Homework Equations


x(t)=Acos(ωt+θ)
v(t)=-Aωsin(ωt+θ)


The Attempt at a Solution



I used the formula, v(t)=-Aωsin(ωt+θ) because you basically have everything you need such as:
A=18.5cm
ω=18.0s^-1
t=0.360s

What I get is:
v(0.360)=-(18.5cm)(18.0)sin(18.0*0.360)
to get -65.1cm/s
which isn't the right answer.
Please tell me where I went wrong!

v=-65.1 cm/s is the velocity. The speed is magnitude of velocity.

ehild
 
  • #5
rude man said:
The highlighted formula makes little sense to me. First, the argument of the cos must be dimensionless but here it looks like it's time.

I believe the formula is just misformatted and is intended to be [itex]\cos(18s^{-1}\times t)[/itex] - so there is no problem with the units.
 
  • #6
Thanks a lot! I got it!
 
Last edited:
  • #7
Borek said:
I believe the formula is just misformatted and is intended to be [itex]\cos(18s^{-1}\times t)[/itex] - so there is no problem with the units.

OK. What can be done to get the OPs to accurately state the problem, I wonder in my oft leisure moments ...
 

FAQ: Simple Harmonic Oscillator Problem

What is a simple harmonic oscillator?

A simple harmonic oscillator is a system that exhibits a repetitive motion around a stable equilibrium point. It can be described by a sinusoidal function and is often used to model the motion of pendulums, springs, and other oscillating systems.

What is the equation for a simple harmonic oscillator?

The equation for a simple harmonic oscillator is x(t) = A cos(ωt + φ), where x is the displacement from equilibrium, A is the amplitude, ω is the angular frequency, and φ is the phase angle.

How do you calculate the period of a simple harmonic oscillator?

The period of a simple harmonic oscillator is the time it takes for one complete cycle of motion. It can be calculated using the equation T = 2π/ω, where T is the period and ω is the angular frequency.

What factors affect the frequency of a simple harmonic oscillator?

The frequency of a simple harmonic oscillator is affected by the mass of the system, the stiffness of the spring (or other restoring force), and the amplitude of the oscillation. It is also dependent on the initial conditions of the system, such as the initial displacement and velocity.

How is a simple harmonic oscillator different from a damped or driven oscillator?

A damped oscillator experiences a decrease in amplitude over time due to the presence of a dissipative force, such as friction. A driven oscillator is one that is forced to oscillate at a specific frequency by an external force. A simple harmonic oscillator, on the other hand, does not experience any external forces or damping and maintains a constant amplitude and frequency of oscillation.

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