What is the Amplitude in an SHM Problem?

In summary, the conversation discussed a problem involving a 245 g mass attached to a horizontal spring oscillating at a frequency of 3.70 Hz. The goal was to determine the amplitude and phase constant using the equations for period, angular frequency, and SHM. The initial conditions provided were that at t=0s, the mass was at x=4.40 cm and had a velocity of -25.0 cm/s. The attempted solution involved using the equation x(t)=Acos(\omegat+\phi) and plugging in t=0 and .0440m for x(0). However, this was incorrect and the phase constant could not be determined without the amplitude. It was suggested to use the equation v
  • #1
jvick717
2
0
Finding amplitude? SHM problem

Homework Statement


A 245 g mass attached to a horizontal spring oscillates at a frequency of 3.70 Hz. At t =0s, the mass is at x= 4.40 cm and has v_x = -25.0 cm/s.
Determine:

Homework Equations


I got the first two
The Period= .27027 s
The Angular Frequency= 23.248 rad/s
The Amplitude=? m rounded to 5 sig figs
The Phase Constant=? rad rounded to 5 sig figs

The Attempt at a Solution



I tried using the equation (x)t=Acos([tex]\omega[/tex]t+[tex]\phi[/tex])
and plugged in t=0 and .0440m for x(0) so the answer i got was .0440m but this is wrong. and i can't attempt the phase constant yet because i need the Amplitude for that. Any help is appreciated thanks.
 
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  • #2


jvick717 said:
I tried using the equation (x)t=Acos([tex]\omega[/tex]t+[tex]\phi[/tex])
and plugged in t=0 and .0440m for x(0) so the answer i got was .0440m but this is wrong. and i can't attempt the phase constant yet
Yes you can. You're off to the right start, but you'll need another equation to determine the phase constant. Hint: there's another value you're given in the problem that you haven't used. What can you do with it?
 
  • #3


I'm guessing that the velocity has something to do with it and maybe the other equation for velocity v(t)= -A[tex]\omega[/tex]sin([tex]\omega[/tex]t+[tex]\phi[/tex])??
 

FAQ: What is the Amplitude in an SHM Problem?

What is amplitude in SHM?

Amplitude in SHM (Simple Harmonic Motion) is the maximum displacement of an object from its equilibrium position during one complete cycle of oscillation. It is a measure of the object's maximum potential energy and is represented by the letter "A".

How do you find the amplitude in an SHM problem?

To find the amplitude in an SHM problem, you need to know the maximum displacement of the object from its equilibrium position during one complete cycle. This can be determined by taking the difference between the maximum and minimum values of the displacement graph or by using the formula A = (xmax - xmin)/2, where xmax and xmin represent the maximum and minimum values of displacement, respectively.

What is the relationship between amplitude and energy in SHM?

The amplitude of an object in SHM is directly proportional to its energy. This means that as the amplitude increases, the object's potential and kinetic energy also increase. Similarly, when the amplitude decreases, the energy decreases as well.

Can the amplitude of an object in SHM ever be negative?

No, the amplitude of an object in SHM is always a positive value as it represents the maximum displacement from the equilibrium position. Negative values would indicate displacement in the opposite direction, but the magnitude of amplitude is always positive.

How does the amplitude affect the period of an object in SHM?

The amplitude does not have a direct effect on the period of an object in SHM. The period is determined by the mass and the spring constant of the object, while the amplitude only affects the object's energy. However, a larger amplitude may result in a longer period due to increased energy and potential for greater resistance from the medium.

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