Trouble finding the amplitude in a SHM problem

AI Thread Summary
The discussion focuses on solving a spring-mass problem in simple harmonic motion (SHM) involving a spring constant of 10.0 N/m and an 80.0 gram mass. The user successfully calculated the angular frequency, linear frequency, and period but struggled with finding the amplitude and phase. Guidance was provided to evaluate the position and velocity equations at t = 0, leading to the conclusion that the maximum speed occurs at the equilibrium position, not at the extended position. The user confirmed that setting the position equation equal to 1.30 cm and using the derivative for velocity would help solve for the amplitude and phase. The conversation emphasizes the importance of understanding the relationship between position, velocity, and energy in SHM problems.
frozen-pizza
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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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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.
 
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?
 
What did you use for vmax? You are not given its value.
 
54.1 cm/s
 
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.
 
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?
 
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.
 
I see thank you so much
 
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