SHO: Find amplitude given k, and x,v,a at unknown time

In summary, the block is experiencing simple harmonic motion and you were given values for position, velocity, and acceleration. You found the mass of the block by setting kx=ma and found the period of oscillation by plugging T=2∏√((kx/a)/k) into the spring period equation. You then found the amplitude by trying to solve for x2 + (v/ω)2, but were lost when you were asked to find it on a vertical plane.
  • #1
jdp1991
6
0

Homework Statement



A block attached to a spring is experiencing simple harmonic motion. You know the value of postion, velocity, and acceleration at an unknown time. Find the period of oscillation, the mass of the block, and the amplitude of oscillation.

We were given values for k, x, v, a

Homework Equations


F=ma=kx
Tspring=2∏√(m/k)
x=Acos(ωt+phi)

The Attempt at a Solution


This was the final question on an exam I just took and I was baffled about how to find Amplitude. We were given values for position, velocity, and acceleration, but I cannot remember what they were off the top of my head, so an algebraic solution is fine.

I started my finding the mass of the block by setting kx=ma: m=kx/a

I then plugged that value for m into the spring period equation: T=2∏√((kx/a)/k)=2∏√(x/a)

The previous two I'm not confident about, but I was completely lost when asked to find the amplitude.

Thanks

EDIT: The problem takes place on a horizontal plane (a flat frictionless surface)
 
Last edited:
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  • #2
How are velocity and acceleration related to the position?
 
  • #3
Is this on a horizontal or vertical plane?
 
  • #4
SammyS said:
How are velocity and acceleration related to the position?

velocity = dx/dt and acceleration = d2x/dt2?
So v = -ωAsin(ωt+phi)
and a = -ω2Acos(ωt+phi)

Villyer said:
Is this on a horizontal or vertical plane?

Sorry horizontal
 
  • #5
So, a/x = -ω2 . Gives you ω.

Then x2 + (v/ω)2 = A2
 
  • #6
SammyS said:
So, a/x = -ω2 . Gives you ω.

ok that makes sense

SammyS said:
Then x2 + (v/ω)2 = A2

maybe I'm missing something obvious, but how did you get this?
 
  • #7
jdp1991 said:
EDIT: The problem takes place on a vertical plane

jdp1991 said:
Sorry horizontal

Still a little lost :3
 
  • #8
Villyer said:
Still a little lost :3

OOPS! Haha sorry it should be horizontal.
 
  • #9
SammyS said:
So, a/x = -ω2 . Gives you ω.

Then x2 + (v/ω)2 = A2

jdp1991 said:
ok that makes sense

maybe I'm missing something obvious, but how did you get this?
Did you try it.

sin2(θ) + cos2(θ) = 1
 

Related to SHO: Find amplitude given k, and x,v,a at unknown time

1. What is SHO and how is it related to finding amplitude given k, x, v, and a at an unknown time?

SHO stands for Simple Harmonic Oscillator, which is a mathematical model used to describe the motion of a mass attached to a spring. In this scenario, the amplitude is the maximum displacement of the mass from its equilibrium position. The values of k (spring constant), x (displacement), v (velocity), and a (acceleration) are all related to the amplitude of the SHO.

2. What is the formula for finding amplitude in SHO?

The formula for finding amplitude in SHO is A = x_max = sqrt((2E)/k), where A is the amplitude, x_max is the maximum displacement of the mass, E is the total energy of the system, and k is the spring constant.

3. Can amplitude be calculated if the time is unknown in SHO?

Yes, amplitude can still be calculated if the time is unknown in SHO. This can be done by using the values of the spring constant, displacement, velocity, and acceleration at a given point in time to determine the total energy of the system, and then plugging that value into the formula for amplitude.

4. How does changing the value of k affect the amplitude in SHO?

The value of k (spring constant) directly affects the amplitude in SHO. A higher value of k will result in a larger amplitude, while a lower value of k will result in a smaller amplitude. This is because a higher spring constant means that the spring is stiffer, making it easier for the mass to be displaced further from its equilibrium position.

5. Can the amplitude in SHO be negative?

No, the amplitude in SHO cannot be negative. The amplitude represents the displacement of the mass from its equilibrium position, so it is always a positive value.

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