What Are the Spring Constant and Mass in This SHM Problem?

In summary, the conversation discusses a problem involving a mass attached to a spring on a frictionless surface. It is compressed by 0.15 m and requires 1.0 J of work to do so. The mass is then released and experiences a maximum acceleration of 14 m/s^2. The conversation also mentions that in order to solve the problem, one must use the fact that the change in potential energy is equal to minus the work done on the mass. The equation Delta U = -W can be used, along with the knowledge that U_{spring} = 1/2kx^2. The conversation ends by suggesting showing progress made on the problem so far for better assistance.
  • #1
foggy
3
0
i need help...

A mass sitting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. 1.0 J of work is required to compress the spring by 0.15 m. If the mass is released from rest with the spring compressed, it experiences a maximum acceleration of 14 m/s^2.

i need to find:
a) The value of the spring constant.
b) The value of the mass.

if anyone can help, i would really appreciate it.
 
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  • #2
Remember that for a mass under the action of a "conservative force" such as the spring force, the change in potential energy is equal to minus the work done on the mass.

Little but important detail: In order to use this fact in your problem, it is necessary to suppose that the mass as been compressed at constant and infinitely small speed, such that we can assert that if the work done by the human force in compressing the mass by 0.15 m is 1.0 J , then the work done by the spring force in compressing the mass by 0.15 m, must be -1.0 J** !

Supposing this condition is met, then the equation

[tex]\Delta U = - W[/tex]

should be helpful, if you remember that [itex]U_{spring}=\frac{1}{2}kx^2[/itex]. As for the part concerning the acceleration, recall how Newton's second law relates force, mass and acceleration.

**You can probably convince yourself of that by considering that under these conditions, the force of spring and the human force differ only in direction, and thus, only is sign!

P.S. You're better off showing us what progress you've made so far in your problems, so we are more able to focus on the points you're having difficulty with.
 
  • #3


Sure, I'd be happy to help you with this simple harmonic motion problem. To start, let's review the formula for the potential energy stored in a spring:

PE = 1/2 * k * x^2

Where PE is the potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position. In this problem, we are given the potential energy (1.0 J) and the displacement (0.15 m). So, we can rearrange the formula to solve for the spring constant:

k = 2 * PE / x^2

Plugging in the values, we get:

k = 2 * 1.0 J / (0.15 m)^2 = 88.89 N/m

So, the value of the spring constant is 88.89 N/m.

Now, let's move on to finding the mass. We can use Newton's second law of motion to relate the maximum acceleration (14 m/s^2) to the mass and the force exerted by the spring:

F = m * a

Where F is the force, m is the mass, and a is the acceleration. In this case, the force is provided by the spring and is equal to the spring constant times the displacement (x):

F = k * x

So, we can rewrite Newton's second law as:

k * x = m * a

Plugging in the values, we get:

88.89 N/m * 0.15 m = m * 14 m/s^2

Solving for m, we get:

m = (88.89 N/m * 0.15 m) / 14 m/s^2 = 0.95 kg

Therefore, the mass is 0.95 kg.

I hope this helps you solve the problem. Remember to always carefully read and understand the given information and use the appropriate formulas to solve for the unknown values. Best of luck!
 

FAQ: What Are the Spring Constant and Mass in This SHM Problem?

What is Simple Harmonic Motion (SHM)?

Simple Harmonic Motion (SHM) is a type of periodic motion where an object oscillates back and forth around an equilibrium point. This motion is characterized by a sinusoidal pattern and is observed in various systems, such as a mass on a spring or a pendulum.

What is the equation for Simple Harmonic Motion?

The equation for Simple Harmonic Motion is given by x = A sin(ωt + φ), where x is the displacement from equilibrium, A is the amplitude (maximum displacement), ω is the angular frequency, and φ is the phase constant.

How do I determine the period of Simple Harmonic Motion?

The period of Simple Harmonic Motion is the time it takes for one complete oscillation or cycle. It can be calculated using the equation T = 2π/ω, where T is the period and ω is the angular frequency.

What factors affect the frequency of Simple Harmonic Motion?

The frequency of Simple Harmonic Motion is affected by the mass, spring constant, and amplitude of the system. A higher mass or spring constant will result in a lower frequency, while a larger amplitude will result in a higher frequency.

How is Simple Harmonic Motion related to energy?

In Simple Harmonic Motion, energy is constantly exchanged between kinetic energy (motion) and potential energy (position). At the equilibrium point, the energy is entirely potential, while at the maximum displacement, the energy is entirely kinetic. This exchange of energy allows the motion to continue indefinitely in the absence of external forces.

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