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## AP®︎/College Physics 1

### Course: AP®︎/College Physics 1>Unit 8

Lesson 2: Simple harmonic motion in spring-mass systems

# Period dependence for mass on spring

David explains what affects the period of a mass on a spring (i.e. mass and spring constant). He also explains what does not affect the period of a mass on a spring (i.e. amplitude and gravitational acceleration). Created by David SantoPietro.

## Want to join the conversation?

• Why isn't dependent on gravity? There must be some tension and hence gravitational acceleration that may affect the period.
• @Eesh Gupta,
It is beyond the scope of a high school physics course, but if you solve the differential equation for a vertical oscillator with the force of gravity included then the solution you obtain includes two additional terms: 1. An additional cosine term with amplitude proportional to the acceleration of gravity but with the same period as the horizontal oscillator (dependent only on the mass and the spring constant), and 2. a constant term equal to the negative amplitude of the additional cosine term; such that the effect of gravity is cancelled out at time = 0. The short answer to your question is that gravity does not affect the period of the oscillator.
• How is the period not dependent on the friction constant?
• It is, but as you may have noted in previous videos in the series, friction has not been considered at all as part of the equations, and such.
• If I double the mass, will it double the period? If I decrease the spring constant by half, would the period double?
• Lets look at the equation:

T = 2π * √(m/k)

If we double the mass, we have to remember that it is under the radical. So this will increase the period by a factor of √2. If we cut the spring constant by half, this still increases whatever is inside the radical by a factor of two. So this also increases the period by √2. Hope this helps!
• If there is no spring and only a normal string it does not depend on mass, why?
• Because heavy things fall with the same acceleration as light things.
• From the equation T=2π√(m/k) , can we say that T² is directly proportional to m and inversely proportional to k?
• Yes, this is an accurate analysis of proportionality.
(1 vote)
• I think David has a mistake: T = 2π√(k/m), not T = 2π√(m/k).

Am I right?
• The equation that is written in the video at about T = 2 π * √(m/k) is correct. As is mentioned in the video increasing the mass increases the period (time it takes for an oscillation) and increasing the spring constant will decrease the period.
• At , David says that the time period isn't dependent on gravity. Then what about the formula for time period which says T= 2*pi* (l/g)^1/2?
• That formula is not valid. The equation should be T = 2 * pi * sqrt(m / k). It is likely that you are thinking of the time period of a gravitational pendulum, which is equal to 2 * pi * sqrt(L / g). This problem is analogous but not equivalent to the spring problem discussed in the video.
(1 vote)
• Hey I tried kinematic equations but it had mass.
Is there any way to calculate the period doesn't depend on the amplitude? I'm still trying to prove/derive it...
• The period does not depend on the Amplitude. The period depends on k and the mass. The more amplitude the more distance to cover but the faster it will cover the distance. The distance and speed will cancel each other out, so the period will remain the same.
• Can someone direct me to the derivation videos mentioned at ? I checked this video: https://www.khanacademy.org/science/physics/mechanical-waves-and-sound/simple-harmonic-motion-with-calculus/v/harmonic-motion-part-2-calculus, but am unclear on how to get from x(t)=Acos(sqrt(k/m)t) to the equation depicted in this video?
• Just curious,
F=ma,
For Simple Harmonic Oscillators,
F=-kx,
Now since x is greater, F is greater
F=m(v-u/t)
Since F is now greater shouldn't t be smaller,
I mean shouldn't they be related?