Vertical SHM and Change in Mass

In summary, a uniform beam with a mass of 225 kg and a 175-kg sack of gravel oscillates in SHM with an amplitude of 40.0 cm and a frequency of .600 cycles/s. When the sack falls off the beam at its maximum upward displacement, the subsequent SHM has a frequency of .8 Hz and an amplitude of .4 m. However, using the incorrect value for angular frequency, the OP got a larger amplitude and energy, which did not make sense. When corrected, the correct frequency and amplitude are found to be .8 Hz and .04 m, respectively. The GPE reference point should also have been chosen better to have the same sign as the displacement.
  • #1
alexfloo
192
0

Homework Statement


A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. The beam has mass 225 kg, and a 175-kg sack of gravel sits on the middle of it. The beam is oscillating in SHM, with an amplitude of 40.0 cm and a frequency of .600 cycles/s.

1.)The sack of gravel falls off the beam when the beam has its maximum upward displacement. What is the frequency of the subsequent SHM of the beam?
2.)What is the amplitude of the subsequent SHM of the beam?
3.)If the gravel instead falls off when the beam has its maximum speed, what is the frequency of the subsequent SHM of the beam?
4.)What is the amplitude of the subsequent SHM of the beam?

Homework Equations


[tex]\omega = \sqrt{\frac{k}{m}}[/tex]
[tex]TE = 1/2kA^2[/tex]

The Attempt at a Solution


I got 1 and 3 pretty easily from the first Equation above. They're both .8 Hz. But I haven't been able to get 2 and 4.

[tex]k = \omega^2 m[/tex]

[tex]k = 144 N/m[/tex]

[tex]TE = 1/2kA^2 = 1/2*144*.04^2 J = .1152 J[/tex]

I'm thinking that at the top of the oscillation, the bag has GPE 2mgA (relative the bottom of oscillation), and so when it falls the energy decreases by that much and amplitude changes accordingly. The only problem is 2*175*9.81*.04 = 137.34 J, much larger than the total energy, which doesn't make sense. Where did I go wrong?
 
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  • #2
Greetings! Try checking your spring constant. Your formula for calculating the spring constant is correct, but it looks like you may have used the wrong value for the angular frequency. Remember that angular frequency equals the frequency (in Hertz) times 2pi.
 
  • #3
Thanks a lot! haha
 
  • #4
I was going over this same problem, could you explain how you got the .8Hz and why you used GPE for the sack to be 2mgh as opposed to the standard mgh?
 
  • #5
also 40 cm is .4 m not .04 m why would you use .4
 
  • #6
EEintraining said:
I was going over this same problem, could you explain how you got the .8Hz and why you used GPE for the sack to be 2mgh as opposed to the standard mgh?

What formula for frequency involves the spring constant and the attached mass?

The OP should have considered placing his GPE reference point at a better location, one that would have the GPE have the same sign as the displacement (signed amplitude).

EEintraining said:
also 40 cm is .4 m not .04 m why would you use .4

Another slip by the OP. Can you correct the problems that have been pointed out and solve the question? Show your work and we can help.
 

1. What is vertical simple harmonic motion (SHM)?

Vertical simple harmonic motion (SHM) is a type of periodic motion in which a body moves up and down along a vertical path due to the force of gravity. The motion is considered simple harmonic if it follows a sinusoidal pattern, with equal displacements on either side of the equilibrium position. This type of motion can occur when a spring or pendulum is attached to a mass and is acted upon by gravity.

2. How is the period of vertical SHM affected by changes in mass?

The period of vertical SHM is not affected by changes in mass. The period is determined by the length of the pendulum or the stiffness of the spring, which are constants in this type of motion. Therefore, increasing or decreasing the mass will not change the period of the motion.

3. What is the relationship between the amplitude of vertical SHM and the change in mass?

The amplitude of vertical SHM is directly proportional to the change in mass. This means that as the mass increases, the amplitude of the motion also increases. This can be seen in the equation for the period of vertical SHM, which includes the mass as a variable.

4. Can the frequency of vertical SHM change if the mass is altered?

No, the frequency of vertical SHM will not change if the mass is altered. The frequency is determined by the period of the motion, which is not affected by changes in mass. However, the amplitude of the motion may change, as discussed in the previous question.

5. How does the change in mass affect the potential energy of a system in vertical SHM?

The potential energy of a system in vertical SHM will increase as the mass increases. This is because the potential energy is directly proportional to the square of the amplitude of the motion, which is affected by changes in mass. Therefore, increasing the mass will result in an increase in potential energy.

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