*Simple frequency and period question - Thanks

[nukeman]

Homework Statement

Ok, below is the question. I am having trouble with getting started. Like the first steps. A and B are the questions I am having trouble with.

"A 1.00 kg glider attached to a spring with a force constant of 25.0 N/m oscillates on a frictionless, horizontal air track. At t = 0, the glider is passing through its equilibrium position with a velocity vx = - 0.150 m/s. (The negative sign means that the glider is headed in the direction which will compress the spring.)

(a) Calculate the frequency and period of the glider’s motion.

(b) Calculate the amplitude of the simple harmonic motion. "

Homework Equations

The Attempt at a Solution

I just don't know how to get start with question A)

EDIT: This is from my notes

Frequency = 1/T
Where T = Period
Or Period = 1/f
Where f is Frequency

But I am not sure how to get the answer to A) for this question from that above data. ?

is this correct for period?

2*pi*sqrt( 1/25) = 1.25

so frequency would be 1 / 1.25 = .8?NOW, is this correct for B) amplitude:

KE = PE, so answer I got was when solving for x was: .03

 

[JHamm]

You should have a formula for the period of a mass on a spring somewhere in your notes.

 

[cupid.callin]

For time you may use time period of SHM
viz $T = 2\pi \Large{\sqrt{\frac{m}{k}}}$

EDIT:

To find this, consider eqn for x coordinate of SHM, $x=Asin(\omega t + \delta)$

This x repeats itself when sin repeats, ie, after 2π or time period T

$sin(\omega t + \delta + 2\pi) = sin(\omega (t+T) + \delta)$

$(\omega t + \delta + 2\pi) = (\omega (t+T) + \delta)$

$T = \Large{\frac{2\pi}{\omega}}$ and in starting of this derivation we assume $\omega = \Large{ \sqrt{\frac{k}{m}} }$

 

[nukeman]

I am very lost on this. Is anything I did correct?

Any help would be fantastic to get me started. Thanks!

 

[asteriatic]

I can provide some guidance on how to approach this problem. First, let's review the equations you have provided from your notes:

Frequency = 1/T

Period = 1/f

Where:

T = period

f = frequency

Now, let's apply these equations to the problem at hand. For question A, we are asked to calculate the frequency and period of the glider's motion. To do this, we need to use the given information about the glider's mass (1.00 kg) and the force constant of the spring (25.0 N/m). We also need to remember that the glider is undergoing simple harmonic motion, which means it is oscillating back and forth with a constant frequency. This frequency is determined by the mass of the glider and the spring constant.

To calculate the frequency, we can use the equation f = 1/2π * √(k/m), where k is the spring constant and m is the mass of the glider. Plugging in the values given in the problem, we get:

f = 1/2π * √(25.0 N/m / 1.00 kg) = 1.26 Hz

Therefore, the frequency of the glider's motion is 1.26 Hz.

To calculate the period, we can use the equation T = 1/f, where T is the period and f is the frequency. Plugging in the frequency we just calculated, we get:

T = 1/1.26 Hz = 0.79 seconds

Therefore, the period of the glider's motion is 0.79 seconds.

Now, for question B, we are asked to calculate the amplitude of the simple harmonic motion. Amplitude is defined as the maximum displacement of the glider from its equilibrium position. In other words, it is the distance between the glider's equilibrium position and its maximum displacement.

To calculate the amplitude, we can use the equation A = x_max, where A is the amplitude and x_max is the maximum displacement. In order to find x_max, we need to use the given information about the glider's initial velocity (vx = -0.150 m/s) and its equilibrium position. We also need to remember that the glider is undergoing simple harmonic motion, which means it is oscillating symmetrically around its equilibrium position. This means that the maximum displacement is equal to the initial