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Class 11 Physics (India)
Course: Class 11 Physics (India) > Unit 18
Lesson 2: Simple harmonic motion in spring-mass systemsSimple harmonic motion in spring-mass systems review
Overview of key terms, equations, and skills for the simple harmonic motion of spring-mass systems, including comparing vertical and horizontal springs.
Equations
| Equation | Symbol breakdown | Meaning in words |
|---|---|---|
| T, start subscript, s, end subscript, equals, 2, pi, square root of, start fraction, m, divided by, k, end fraction, end square root | T, start subscript, s, end subscript is the period of the spring, m is the mass, and k is the spring constant. | The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant. |
How to analyze vertical and horizontal spring-mass systems
Both vertical and horizontal spring-mass systems without friction oscillate identically around an equilibrium position if their masses and springs are the same.
For vertical springs however, we need to remember that gravity stretches or compresses the spring beyond its natural length to the equilibrium position. After we find the displaced position, we can set that as y, equals, 0 and treat the vertical spring just as we would a horizontal spring. Figure 1 below shows the resting position of a vertical spring and the equilibrium position of the spring-mass system after it has stretched a distance d.
We can use a free body diagram to analyze the vertical motion of a spring mass system. We would represent the forces on the block in figure 1 as follows:
Then, we can use Newton's second law to write an equation for the net force on the block:
The block in figure 1 is not accelerating, so our equation simplifies to:
Common mistakes and misconceptions
Sometimes people think that the period of a spring-mass oscillator depends on the amplitude. Increasing the amplitude means the mass travels more distance for one cycle. However, increasing the amplitude also increases the restoring force. The increase in force proportionally increases the acceleration of the mass, so the mass moves through a greater distance in the same amount of time. Thus, increasing the amplitude has no net effect on the period of the oscillation.
Learn more
For deeper explanations of spring-mass systems see the video on period dependence for a mass on a spring.
To check your understanding and work toward mastering these concepts, check out our exercises:
Want to join the conversation?
Say we have a horizontal SHM frictionless spring-mass system. If I take another mass and drop it on top of the mass in oscillation, will it change the amplitude? I did a specific example and solved for the new amplitude with the new velocity after the mass was added and got the same amplitude. Am I correct to get the same amplitude?(7 votes)
you actually are correct because by adding more mass you are simply increasing the force and hence the acceleration of mass moving through the distance in the same amount of time. the distance ( amplitude ) doesn't change because more mass was added, cause mass has no effect on it.
more mass means more force= more acceleration moving through a greater distance in the same amount of time.(3 votes)
I actually derived the formula of k = 4π^2m/T^2 by differentiating the sin(t) function of displacement twice to find the acceleration, then multiply by mass and divide by amplitude to find spring constant.
First by finding the specific sin(t) function in the form of Asin(Bt), through the given amplitude(A) and period(T).
1. Knowing that B•T = 2π. Hence B = 2π/T. And A is just the Amplitude. Giving the Asin(Bt) equation as (Amplitude)sin(2πt/T)
2. Deriving once will give (A2π/T)sin(2πt/T), which is the function of velocity.
3. Deriving again will give -(A4π^2/T^2)sin(2πt/T)
4. Knowing that at k = (acceleration•mass)/displacement. Since we know the amplitude, and we also know the maximum displacement is at t = T/4, which is at Bt = π/2, which is when sin(Bt) = 1, simultaneously having the greatest acceleration of this oscillating system.
5. So we can get maximum acceleration = -A4π^2/T^2. Hence force = Ma, force = -A4Mπ^2/T^2. And k = Force divide by displacement, giving k = maximum force divide by the simultaneous maximum displacement, which gives k = (-A4Mπ^2/T^2)/A.
6. So the A cancels out, giving k = -4Mπ^2/T^2. This negative sign means the direction of force is inwards, which can be neglected. k = 4Mπ^2/T^2(3 votes)
would changing the mass (and nothing else) of an object change the graph of simple harmonic motion?(2 votes)
Yes-changing the mass would change the time taken to complete one full cycle (the period) and therefore would change the graph.(2 votes)
- So, why does loaded spring oscillation is s.h.m? (simple harmonic motion)(2 votes)
- So for a mass undergoing SHM, where is the mass, when the force on it has its lowest magnitude?(1 vote)
- Explain what will happen to the periodic time? if you change the spring constant to the low value.(1 vote)
- If I take a spring-mass system to 0 gravity , what would be the periodicity??(1 vote)
It would remain the same everywhere in the universe as the time period is independent of acceleration by gravity.(1 vote)
- which if the following is/are characteristics of simple harmonic motion?
I. The acceleration is constant
II. the frequency is independent of the amplitude
III. it requires a restoring force that point in a fixed direction(1 vote) 
A 0.5 kg mass hangs on a spring, find it’s period
of oscillation(0 votes)
For a mass attached to a spring, the period of oscillation is equal to 2π √(m/k). Plug in 0.5 for m and if you know what the spring constant k is you can solve.(2 votes)
- A 125 N object vibrates with a period of 3.56s when hanging from a spring. What is the spring constant of the spring?(0 votes)
F=ma, (a=g=9.8 m/s/s)
find m, substitute in
k=m*4*pi^2/T^2
voila, you get ur answer,
k = 39.735 N/m(2 votes)