SHM acceleration problem

In summary, a 29.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x=.350 m, then let's go. The mass undergoes simple harmonic motion with a period of 3.50 s. The position of the mass 2.975 s after release is .206 m. The magnitude of the maximum acceleration the mass undergoes during its motion is not correctly calculated using the equation ma=-kx. To find the maximum acceleration, one must consider the equation for acceleration and the relationship between force and acceleration.
  • #1
A 29.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x=.350 m, then let's go. The mass undergoes simple harmonic motion with a period of 3.50 s. What is the position of the mass 2.975 s after the mass is released?
I got this part fine, with an answer of .206 m
Consider the same mass and spring discussed in the previous problem. What is the magnitude of the maximum acceleration the mass undergoes during its motion?

I found the angular velocity by using
[tex]\omega = 2\pi/ T [/tex] and got it to be 1.795. I then plugged it into the equation, [tex] a= -\omega^2 x_{m}cos(\omega t +\theta) [/tex]
with x= .206 m, and t=2.975 s. I got a= .389 m/S^2, which isn't right.
can someone tell me what I'm doing wrong?
 
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  • #2
Punchlinegirl said:
A 29.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x=.350 m, then let's go. The mass undergoes simple harmonic motion with a period of 3.50 s. What is the position of the mass 2.975 s after the mass is released?
I got this part fine, with an answer of .206 m
Consider the same mass and spring discussed in the previous problem. What is the magnitude of the maximum acceleration the mass undergoes during its motion?

I found the angular velocity by using
[tex]\omega = 2\pi/ T [/tex] and got it to be 1.795. I then plugged it into the equation, [tex] a= -\omega^2 x_{m}cos(\omega t +\theta) [/tex]
with x= .206 m, and t=2.975 s. I got a= .389 m/S^2, which isn't right.
can someone tell me what I'm doing wrong?

You were asked for the maximum acceleration. That is not what you calculated. Where is acceleration maximum, or what is the maximum based on your acceleration equation
 
  • #3
Ok I used the ma=-kx. Solving for k with my angular velocity gave k=93.4 N/m. Plugging in m= 29 kg and x= .206 m gave a=.664 m/s^2. This wasn't right. I understand that I wasn't solving for max acceleration before, but I don't think I know an equation to do it.
 
  • #4
Punchlinegirl said:
Ok I used the ma=-kx. Solving for k with my angular velocity gave k=93.4 N/m. Plugging in m= 29 kg and x= .206 m gave a=.664 m/s^2. This wasn't right. I understand that I wasn't solving for max acceleration before, but I don't think I know an equation to do it.

Look carefully at the equation that you wrote for the acceleration. Also, think about when the force is greatest and use what you know about the relationship between force and acceleration. Either of these (you don't need both, but they are connected) will get you to the answer.
 

1. What is SHM acceleration problem?

The SHM acceleration problem is a physics problem that involves calculating the acceleration of an object undergoing simple harmonic motion (SHM). This type of motion is characterized by a periodic back-and-forth motion, where the acceleration is directly proportional to the displacement from the equilibrium position.

2. What is the equation for SHM acceleration?

The equation for SHM acceleration is a = -ω²x, where a is the acceleration, ω is the angular frequency, and x is the displacement from equilibrium position. This equation is derived from Hooke's Law and Newton's Second Law of Motion.

3. How do you solve SHM acceleration problems?

To solve SHM acceleration problems, you need to follow a few steps. First, identify the given variables and their values. Then, use the equation a = -ω²x to calculate the acceleration. Finally, plug in the values for the given variables to solve for the unknown variable.

4. What is the difference between SHM acceleration and SHM velocity?

SHM acceleration and SHM velocity are two related but different concepts. SHM acceleration refers to the rate of change of velocity with respect to time, while SHM velocity refers to the rate of change of displacement with respect to time. In other words, acceleration measures how quickly the velocity is changing, while velocity measures how quickly the position is changing.

5. What are some real-life examples of SHM acceleration?

SHM acceleration can be observed in many real-life situations, such as the motion of a pendulum, the vibration of a guitar string, or the oscillation of a diving board. It is also commonly seen in mechanical systems, such as springs and shock absorbers, where the system is designed to undergo SHM to reduce the effects of external forces.

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