Trouble finding the amplitude in a SHM problem

In summary, In part a, a horizontal spring-mass is found to have a velocity of 54.1 cm/s in the negative x direction. In part b, the amplitude of the mass is found to be. .541 and the frequency is found to be 11.2 Hz.
  • #1
frozen-pizza
7
0

Homework Statement


a horizontal spring-mass is composed of a spring with constant 10.0 N/m and an 80.0 gram mass on the end of the spring. the surface supporting the mass is friction less. when the system is first observed, the spring is extended 1.30 cm and the velocity of the mass is 54.1 cm/s in the negative x direction
a. find angular frequency, linear frequency, and the period.
b. apply the given info and your results from part a to find the values for Xsub m, and Φsub m.

Homework Equations


w=sqrt(k/m), f= w/2pi, and t=1/f
x=Acos(w+Φ)

The Attempt at a Solution


i found w=11.2 f=1.78 and t=.56
for part b i seem to be totally stuck, i would really appreciate a push towards the right direction.
i looking at conservation of energy to find the amplitude but i don't know what energy is so i get stuck with two unknowns.
 
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  • #2
Hi frozen-pizza and welcome to PF.
You don't need energy. Just evaluate x(t) and dx/dt at t = 0 and set them equal to the given values.
 
  • #3
for amplitude i used Vmax=Aw so A=.541/11.2 and i seem to get the right answer, is this right or just a coincidence?
 
  • #4
What did you use for vmax? You are not given its value.
 
  • #5
54.1 cm/s
 
  • #6
That's the velocity when the spring is at position 1.30 cm. The speed is maximum when the spring is unstretched at the equilibrium position.
 
  • #7
yea that makes sense, it just happened to work out then.
ok so if i set x(t)=0 is x(t)=Acos(0+Φ)? if so do i have to set it equal to 1.30 cm?
 
  • #8
frozen-pizza said:
k so if i set x(t)=0 is x(t)=Acos(0+Φ)? if so do i have to set it equal to 1.30 cm?
Yes. Then evaluate the derivative at t = 0 and set it equal to -54.1 cm/s. Solve the system of two equations and two unknowns.
 
  • #9
I see thank you so much
 

1. What is SHM and why is it important to find the amplitude?

SHM stands for Simple Harmonic Motion, which is a type of oscillatory motion where the restoring force is directly proportional to the displacement from the equilibrium position. Finding the amplitude in a SHM problem is important because it is a measure of the maximum displacement from the equilibrium position and can help determine the energy and period of the oscillation.

2. How do I know if a problem involves SHM?

A problem involves SHM if it meets the following criteria: the restoring force is directly proportional to the displacement, the motion is periodic, and the motion is along a single axis. Some examples of SHM include a mass on a spring, a simple pendulum, and a vibrating guitar string.

3. What is the formula for calculating the amplitude in a SHM problem?

The formula for calculating the amplitude is A = xmax - xeq, where A is the amplitude, xmax is the maximum displacement from the equilibrium position, and xeq is the equilibrium position. This formula can be used for any type of SHM problem, as long as the equilibrium position is known.

4. Are there any tricks or shortcuts for finding the amplitude in a SHM problem?

One trick for finding the amplitude in a SHM problem is to use the Pythagorean theorem. If the motion is along a single axis, the amplitude can be calculated using the maximum displacement in the x-direction and the maximum displacement in the y-direction. Another shortcut is to use the formula A = xmax/2, which works for a mass on a spring with a maximum displacement of xmax.

5. Can the amplitude in a SHM problem ever be negative?

No, the amplitude in a SHM problem can never be negative. It is a measure of the distance from the equilibrium position, so it will always be a positive value. If the amplitude is given as a negative value, it is likely that the problem is referring to the displacement from the equilibrium position in the opposite direction.

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