Simple Harmonic Motion - Mass and massive spring

In summary: The cosine and sine of maximum displacement and maximum speed are the opposite of each other. So, at maximum displacement the cosine is negative, and at maximum speed the sine is positive.
  • #1
DaviBones
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0

Homework Statement



A block of mass M is connected to a spring of mass m and oscillates in simple harmonic motion on a horizontal, frictionless track (Figure P15.62). The force constant of the spring is k and the equilibrium length is l. Assume that all portions of the spring oscillate in phase and that the velocity of a segment dx is proportional to the distance x from the fixed end; that is,

[itex]v_{x}=\frac{x}{l}v[/itex]

Also, note that the mass of a segment of the spring is

[itex]dm=\frac{m}{l}dx[/itex]

(a) Find the kinetic energy of the system when the block has a speed v, using m, M, and v as necessary.

(b) Find the period of oscillation, using m, M, and k as necessary.


Homework Equations



The above two equations. Also,

[itex]K=\frac{1}{2} \int{v^{2}}dm[/itex]

[itex]v=-A \sqrt{\frac{k}{m}} sin(t \sqrt{\frac{k}{m}})[/itex]

[itex]T=2 \pi \sqrt{\frac{m}{k}}[/itex]


The Attempt at a Solution



(a) I simply plugged and chugged using the two given equations and the kinetic energy equation. The only tricky part was realizing that when the velocity and kinetic energy are at their maximum, x=l , meaning, the distance the from the wall to the mass is the same as the distance from the wall to the equilibrium point.

(b) Not so simple. I can't find any equations for period other than the one above, and that one isn't really helping me. I could just plug in M and m into it and add the resulting equations, but I have a feeling the term with m in it would have some sort of denominator, like the term with m in my correct equation for kinetic energy.

Suggestions? Is my intuition failing me, or is there something I'm missing?
 
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  • #2
Concerning the vibration of the spring, there are two variables. One is the position x of a little piece of mass dm along the length of the unstretched spring, and the other variable is Δx, the displacement of that piece of string from its equilibrium position. All pieces move in phase, with velocity v(x)=x/L V: V is the velocity of the object of mas M at the end of the spring. The equilibrium position of that object is X=L, and its displacement and velocity are ΔL=Acos(ωt), V=-Aωsin(ω).

Write out the KE of a small piece of string as function of x, than integrate for the whole length of the spring to get the KE of the whole spring. Add the KE of the object of mass M.
The potential energy (elastic energy) is that of the spring, of spring constant k and stretched by ΔL. KE+PE= const, so the kinetic energy at the equilibrium position is the same as the potential energy of the spring when ΔL is at maximum. You get the angular frequency ω from this condition.

ehild
 
  • #3
Ahahaha! I got it. Man, physics/math at it's finest. Thanks, those last two sentences really helped me out. Once I figured out how to get rid of the sine/cosine terms in the conservation of energy terms, it was a piece of cake.
 
  • #4
What is cosine/sine at maximum displacement/maximum speed?

ehild
 
Last edited:
  • #5


Your solution for part (a) seems correct. For part (b), you can use the equation T=2π√(m/k), which is the correct equation for the period of a mass-spring system in simple harmonic motion. However, in this case, the mass (m) is not just the mass of the block, but also includes the mass of the spring (m) as given in the problem. So the equation should be T=2π√((m+M)/k). This accounts for the fact that the spring also has mass and affects the period of the oscillation.
 

1. What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction of the displacement. This results in a back-and-forth motion around the equilibrium point.

2. How is SHM related to mass and massive spring?

In SHM, the mass and stiffness of a spring are the two main factors that determine the frequency and period of the oscillation. A heavier mass and a stiffer spring will result in a slower frequency and longer period, while a lighter mass and a more flexible spring will result in a faster frequency and shorter period.

3. What is the equation for calculating the period of SHM?

The period (T) of SHM can be calculated using the formula T = 2π√(m/k), where m is the mass of the object and k is the spring constant.

4. How does amplitude affect SHM?

The amplitude of SHM is the maximum displacement from equilibrium. It does not affect the period or frequency of the oscillation, but it does determine the maximum potential energy and the maximum velocity of the object in motion.

5. Can SHM occur without a spring?

Yes, SHM can occur without a spring. It can also occur in other systems, such as a simple pendulum or a mass attached to a string. The key is that there must be a restoring force that is directly proportional to the displacement from equilibrium.

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