Simple Harmonic Motion of a Spring?

In summary, if a mass is attached to a spring and has simple harmonic motion, then the acceleration is given by the equation X(t) = 5(sin(3pi(t))).
  • #1
EthanVandals
55
2

Homework Statement


If a mass attached to a spring has motion given by the equation X(t) = 5(sin(3pi(t))), what is the equation for the acceleration of the spring? What is the angular speed of the spring,and what is its frequency and period? If the spring has a spring constant of 900 N/m, what is the mass attached at the end of the spring. Assume that the spring is constant, frictionless, and exhibits simple harmonic motion.

Homework Equations

The Attempt at a Solution


I don't even know where to begin on this one. I looked through the slides and nowhere is there a formula for acceleration that doesn't require you to know the mass already. I feel like I'm either missing an equation or missing a variable here that is required to solve this problem. Thanks in advance!
 
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  • #2
The acceleration is the second derivative of the position. That's all you need for starters.
 
  • #3
kuruman said:
The acceleration is the second derivative of the position. That's all you need for starters.
Is there a way to do it without taking the derivative? I could do that, but this course is supposed to be solely algebra based...
 
  • #4
EthanVandals said:
Is there a way to do it without taking the derivative? I could do that, but this course is supposed to be solely algebra based...
Do you mean it is not supposed to involve calculus?
 
  • #5
haruspex said:
Do you mean it is not supposed to involve calculus?
Yes. The professor mentioned that calculus is a method of solving these kinds of problems and many others, but he said we are not supposed to use calculus whatsoever.
 
  • #6
EthanVandals said:
Yes. The professor mentioned that calculus is a method of solving these kinds of problems and many others, but he said we are not supposed to use calculus whatsoever.
The only alternative I can think of for this problem is to recognise that the equation represents simple harmonic motion and quote the acceleration etc. equations that go with it. That is also suggested by this comment:
EthanVandals said:
Assume that the spring ... exhibits simple harmonic motion.
But calculus is used to obtain those equations, so while you do not write any calculus yourself it is not really avoiding the use of calculus.
Seems a pointless ordinance to me.
 
  • #7
haruspex said:
The only alternative I can think of for this problem is to recognise that the equation represents simple harmonic motion and quote the acceleration etc. equations that go with it. That is also suggested by this comment:

But calculus is used to obtain those equations, so while you do not write any calculus yourself it is not really avoiding the use of calculus.
Seems a pointless ordinance to me.
I'm glad I'm not the only one who thinks so. Thank you for the advice!
 
  • #8
EthanVandals said:

Homework Statement


If a mass attached to a spring has motion given by the equation X(t) = 5(sin(3pi(t))), what is the equation for the acceleration of the spring? What is the angular speed of the spring,and what is its frequency and period? If the spring has a spring constant of 900 N/m, what is the mass attached at the end of the spring. Assume that the spring is constant, frictionless, and exhibits simple harmonic motion.

Homework Equations

The Attempt at a Solution


I don't even know where to begin on this one. I looked through the slides and nowhere is there a formula for acceleration that doesn't require you to know the mass already. I feel like I'm either missing an equation or missing a variable here that is required to solve this problem. Thanks in advance!
upload_2017-5-9_19-58-18.png
, in this case your value of omega is 3pi so you could equate the two and solve for whatever you need.
 
  • #9
EthanVandals said:
Yes. The professor mentioned that calculus is a method of solving these kinds of problems and many others, but he said we are not supposed to use calculus whatsoever.
OK then. This problem was not created from a vacuum. Your professor must have taught you something about simple harmonic motion and given you some equations. How about listing the relevant equations? Look at your notes and the textbook.
 
  • #10
haruspex said:
But calculus is used to obtain those equations, so while you do not write any calculus yourself it is not really avoiding the use of calculus.
Seems a pointless ordinance to me.
Yes, indeed calculus is used to obtain these equations, however the ordinance may not be as pointless as it seems, but designed to test whether one understands the basics of simple harmonic motion.

In simple harmonic motion, the basic idea is that the acceleration is always proportional to the displacement from equilibrium, the constant of proportionality being ##-\omega^2##. One does not need calculus to derive this result for a spring-mass system; it follows from ##F_{net} = ma = -kx## and the definition ##\omega^2 = k/m##. Thus, if one is given an expression for the displacement as a function of time, one can extract ##\omega## from this expression (see post #8) and then just write down the acceleration as a function of time. The rest of the questions may be answered from a knowledge of ##\omega##.
 
  • #11
kuruman said:
In simple harmonic motion, the basic idea is that the acceleration is always proportional to the displacement from equilibrium, the constant of proportionality being ##-\omega^2##. One does not need calculus to derive this result for a spring-mass system; it follows from ##F_{net} = ma = -kx## and the definition ##\omega^2 = k/m##. Thus, if one is given an expression for the displacement as a function of time, one can extract ##\omega## from this expression (see post #8) and then just write down the acceleration as a function of time. The rest of the questions may be answered from a knowledge of ##\omega##.
Yes, I realized that without calculus you could get as far as a = -kx/m, but you have no basis for replacing k/m with the square of the given angular frequency, 3π. So you still cannot answer the first question.
I agree it is not completely pointless - it is an exercise in remembering and applying some standard equations. I'm a bit biased because I never was much good at remembering equations, but I could always remember how to derive them.
 
  • #12
haruspex said:
Yes, I realized that without calculus you could get as far as a = -kx/m, but you have no basis for replacing k/m with the square of the given angular frequency, 3π. So you still cannot answer the first question.
I agree it is not completely pointless - it is an exercise in remembering and applying some standard equations. I'm a bit biased because I never was much good at remembering equations, but I could always remember how to derive them.
I tried what they said above and I was able to get some answers. Probably not enough to get an A, but maybe enough to get some points to add to the other problems. I do not think you are biased. The suggestions about using my notes and textbook will not work, as there were no notes involving equations of SHM and there is no textbook, and I'm not quite smart enough to understand it from looking it up. :/ thanks for trying to help however!
 

Related to Simple Harmonic Motion of a Spring?

1. What is Simple Harmonic Motion?

Simple Harmonic Motion refers to the back and forth motion of an object that occurs when the restoring force is proportional to the displacement from the equilibrium position. This type of motion is commonly observed in systems such as springs, pendulums, and mass-spring systems.

2. What is a spring constant?

A spring constant, also known as the force constant, is a measure of the stiffness of a spring. It is represented by the letter "k" and is the ratio of the force applied to the displacement caused by the force. In simple harmonic motion, the spring constant determines the frequency and period of the motion.

3. How is the period of a spring's motion calculated?

The period of a spring's motion can be calculated using the formula T = 2π√(m/k), where T is the period in seconds, m is the mass attached to the spring in kilograms, and k is the spring constant in newtons per meter. This formula assumes that the amplitude of the motion is small and that the spring is ideal.

4. What factors affect the frequency of a spring's motion?

The frequency of a spring's motion is affected by the mass attached to the spring, the spring constant, and the amplitude of the motion. Increasing the mass or the spring constant will decrease the frequency, while increasing the amplitude will increase the frequency.

5. How is energy conserved in simple harmonic motion of a spring?

In a simple harmonic motion of a spring, energy is conserved because the total mechanical energy (potential energy + kinetic energy) remains constant throughout the motion. As the spring compresses or stretches, the potential energy increases, while the kinetic energy decreases. However, the total energy remains the same, leading to a continuous back and forth motion.

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