How to find period of a SHM concerning a cube and spring?

In summary: So for rotational oscillation, you would use the small angle approximation for sin(ωt) and then use the solution for ω to solve for θ.
  • #1
Ari
3
0
1. Homework Statement
The 4.00 kg cube in the figure has edge lengths d = 8.00 cm and is mounted on an axle through its center. A spring ( k = 1400 N/m ) connects the cube's upper corner to a rigid wall. Initially the spring is at its rest length. If the cube is rotated 4.00° and released, what is the period of the resulting SHM?

bvCDWNW.png


m = 4 kg
d = .08 m
k = 1400 N/m
Θ = 4°
T = ?

2. Homework Equations
I = (1/6)Md2
F = ma
F = -kx = -(mω2x)
k = mω2
τ = -d(FgsinΘ)

3. The Attempt at a Solution
Restoring force is:
F = -kx while x = dsinΘ so...
F = -kdsinΘ
F = -(1400)(.08)sin(4°)
F = -7.8127 N

Newton's 2nd Law gives:
F = ma = F/m
a = (-7.8127)/4 = -1.9532

Using k = mω:
k = mω
ω = k/m = 1400/4 = 350

Using F = -kx:
x = F/-k
x = (-7.8127)/(-1400) = .00558 m

Restoring torque is:
τ = -dmgsinΘ while τ = Iα so...
Iα = -dmgsinΘ
α = -(.08)(4)(9.81)sin(4°) = -.219

Rotational to linear:
a = αr , r = .1131 m
a = -.0248

I'm not sure where to go with this further, and to be honest, not even sure if I'm going in the right direction.
Another thing is I'm not sure whether I'm supposed to use T = 2π √(m/k) or another equation, since that equation is for SHM of springs, but I can't seem to get anywhere with it.
 

Attachments

  • bvCDWNW.png
    bvCDWNW.png
    994 bytes · Views: 1,469
Physics news on Phys.org
  • #2
The equation of motion for a mass suspended by a spring is a = d^2 y/dt^2 = (k/m) y which leads to a period, T = 2 pi sqrt(m/k) What you want to set up is an equation for alpha (the angular acceleration) (the second derivative of theta with respect to time) in terms of theta. You can then find the period in terms of the coefficient of theta analogous to the simple spring-mass problem.
 
  • #3
Ari said:
x = dsinΘ
Check that.
Ari said:
the cube's upper corner
That does not match the picture, which shows one edge uppermost, not one corner. I would guess the diagram is as intended.
Ari said:
If the cube is rotated 4.00°
Is this relevant? In SHM, does the frequency depend on the amplitude?
Ari said:
Using k = mω:
That is for a mass moving linearly on the end of a spring. It does not apply here. What kind of motion is this? Do you know any frequency formula for this kind of motion? If not, can you derive the differential equation for the motion?
 
  • #4
haruspex said:
Check that.

That does not match the picture, which shows one edge uppermost, not one corner. I would guess the diagram is as intended.

Is this relevant? In SHM, does the frequency depend on the amplitude?

That is for a mass moving linearly on the end of a spring. It does not apply here. What kind of motion is this? Do you know any frequency formula for this kind of motion? If not, can you derive the differential equation for the motion?

This would be a rotational motion I believe, since you are applying a force that causes the cube to turn, which would be angular frequency ω.
The only equation I know relating to angular frequency ω would be ω = 2π/T = 2πf.

And for the diagram, the cube's upper corner would be concerning the corner attached to the spring I believe.
 
  • #5
Ari said:
The only equation I know relating to angular frequency ω would be ω = 2π/T = 2πf
Need to be careful with use of ω in rotational oscillation. The displacement is an angle,θ say. It is common to write ω for ##\dot\theta## in other contexts, but this is different from the ω in the sin(ωt) in SHM, so better not to do that here.
θ is analogous to the x in linear oscillation, so ##\dot\theta## is like ##\dot x##. Stick to the ##\dot\theta## and ##\ddot\theta## notation.
Think about torque and angular acceleration and try to write the differential equation for θ.*
Ari said:
And for the diagram, the cube's upper corner would be concerning the corner attached to the spring I believe.
Yes, but the diagram does not show an upper corner of a cube so attached, it shows the middle of one edge attached, or maybe a corner of a square lamina. Suspend a cube by one corner and view it from the side. Does it look like the diagram?

*Edit: Actually, you almost had it in post #1, but you made the mistake of plugging in the initial angle instead of leaving it as a general equation relating θ to α (##=\ddot\theta##).
And as Dr Dr news posts, you need to use the small angle approximation for sine. You have come across that with pendulums, I would think.
 
Last edited:
  • #6
The point is that having solved one differential equation say d^2 x/dt^2 = -(k/m) x, x = A sin ϖt, ω = sqrt(k/m) Any other differential equation of the same form say d^ θ/dt^2 = -(τ/I) θ will have a similar solution. θ = A sin ωt , ϖ = sqrt(τ/I) , τ is the restoring torque and I is the moment of inertia. one other consideration, the restoring torque is proportional to sin θ which for small θ, can be approximated by θ.
 
Last edited:

Related to How to find period of a SHM concerning a cube and spring?

1. What is SHM and how does it relate to a cube and spring?

SHM stands for Simple Harmonic Motion, which is a type of motion where an object oscillates back and forth between two points. In the case of a cube and spring system, the cube is attached to the spring and moves back and forth due to the restoring force of the spring.

2. How can I determine the period of a SHM for a cube and spring?

The period of a SHM is the time it takes for one complete oscillation. For a cube and spring system, the period can be found using the equation T = 2π√(m/k), where T is the period, m is the mass of the cube, and k is the spring constant.

3. What factors affect the period of a SHM for a cube and spring?

The period of a SHM for a cube and spring is affected by the mass of the cube, the spring constant of the spring, and the amplitude of the oscillation. As the mass or spring constant increase, the period will also increase. On the other hand, as the amplitude increases, the period will decrease.

4. Can the period of a SHM for a cube and spring be changed?

Yes, the period of a SHM for a cube and spring can be changed by altering the factors that affect it. For example, by changing the mass or spring constant, the period can be increased or decreased. Additionally, changing the amplitude of the oscillation will also affect the period.

5. Are there any real-life applications of a cube and spring SHM system?

Yes, there are many real-life applications of a cube and spring SHM system. Some examples include the suspension system in a car, the motion of a pendulum, and the movement of a tuning fork. Understanding the period of a SHM for a cube and spring can also be helpful in designing and analyzing mechanical systems.

Similar threads

  • Introductory Physics Homework Help
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
927
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
10K
Back
Top