# Simple harmonic motion glider problem

In summary: I think adrian is looking for a way to use more calculus. Hence I didn't want to give the answer away. But yes, your way is simpler.In summary, a glider attached to a spring is released from rest and oscillates with a period of 2.40 s and a maximum speed of 46.0 cm/s. By using the equations x(t) = A cos (wt + q) and v(t) = -Aw sin (wt), the amplitude of the oscillation is found to be 17.57 cm. It is also noted that the maximum speed occurs at the equilibrium point of the motion.
An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t = 0 s . It then oscillates with a period of 2.40 s and a maximum speed of 46.0 cm/s.

What is the amplitude of the oscillation?

i tryed solving for t by using the equation x(t) = A cos (wt + q) with w = angular frequency and q = phase constant A = amplitude, i tried to solve for the 3 variables by using the equatoin, the derivitive of the equation, and the 2nd derivitive of the equation ... unsuccesful.

is there an equation other than that one that allows me to calculate the simple harmonic motion? please tell me how to do this problem...

You know some initial conditions. If it is released from rest, then x'(0) = 0

You also know that it is released from rest after being stretched to its maximum displacement: so x(0) = A.

plugging in t = 0

Acos(w(0) + q) = Acos(q) = A

cos(q) = 1

q = 0

You have yet another bit of info: what does the maximum speed tell you (at what point in SHM is the oscillator moving fastest?)

i see what you're saying

the maximum speed occurs when it is at its equlibrium point, x = 0 right?
using the first derivitive i get velocity equation, which is:
v(t) = - A w sin (wt) w = 2.61
-46 = - A*2.61 sin(2.61*t) <---this is at the eqi. point, going to the left

but i still have 2 variables in this equation, and i can't seem to solve it with
x(t) = A cos (wt)
0 = A cos (2.61 t) <---this is at the eqi. point, going to the left

how do i solve this set of equations:
-46 = - A*2.61 sin(2.61*t)
0 = A*cos (2.61*t)

Your equations are not quite right, because you haven't figured out at what time they are true. You are still using the variable t. Why? The equations are not true for all time. They are only true at a specific instant (when the object is in the middle). At what time is that? How would you figure it out?

Hint1: you haven't made use of the information given about the period of the motion yet (and by that I mean, you haven't made use of it directly).

Hint2: The bottom equation, even when corrected, will not be required, for it will essentially contain the information: 0=0, at the right time (when the object is at the equilibrium position).

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Look carefully at the equation for v. When is v maximized?

Nah! For what purpose? You should already know when v is at its maximum in SHM. That's what my hint1 was alluding to. Don't mess him up! Just follow my hints, adrian.

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cepheid said:

Can you explain what you mean by that? :( Isn't it the amplitude that he is looking for, given the period of oscillation?

waiiiit a second..period 2.4 s that's going forward and back...
cut that by half, that's going one way trip
cut that by half again...would it be..could it be..AT THE EQILIBRIUM POINT!?
2.4/4 = 0.6

so the time to arrive at the equilibrium point is...0.6!

now i see, the bottom equation does reduce to 0 = 0 when i plug in the t in

-46 = - A*2.61 sin(2.61*t) t = 0.6
-46 = - A*2.61 sin(2.61*0.6)
A = 17.57!

SWEETTT thanks for all your help!

ramollari said:
Can you explain what you mean by that? :( Isn't it the amplitude that he is looking for, given the period of oscillation?

Ramollari: I edited it for clarity, to explain what I meant. Didn't mean to be rude. And adrian figured it out too, illustrating the point perfectly.

I know you were saying it was at a maximum when sinwT = 1,

wT = pi/2

solve for T and you get the same answer, and equally valid and more mathematical way of looking at it, so sorry I jumped on you.

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This is how I'd go to solve it:

$$|v_{max}| = \omega A = \frac{2 \pi}{T}A$$

I would just solve for A

$$A = \frac{v_{max}T}{2\pi} = 17.57cm/s$$.

I think this is more elegant, though the way you solved it is also good.

lol yes, I keep editing my posts on you, which is bad. Yes, as I stated above, I agree it is more elegant.

## 1. What is simple harmonic motion?

Simple harmonic motion is a type of oscillatory motion in which a system moves back and forth around a central equilibrium point, with a restoring force that is directly proportional to the displacement from the equilibrium point.

## 2. How does a glider experience simple harmonic motion?

A glider experiences simple harmonic motion when it is attached to a spring or elastic material and is allowed to move freely back and forth along a horizontal track.

## 3. What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x(t) = A*cos(ωt + φ) where x is the displacement, A is the amplitude, ω is the angular frequency, and φ is the phase constant.

## 4. How does the mass of the glider affect its simple harmonic motion?

The mass of the glider does not affect the frequency of its simple harmonic motion, but it does affect the amplitude. A heavier glider will have a smaller amplitude and vice versa.

## 5. What is the period of a glider in simple harmonic motion?

The period of a glider in simple harmonic motion is the time it takes for the glider to complete one full oscillation, which is equal to 2π/ω where ω is the angular frequency.

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