Vertical SHM and Change in Mass

AI Thread Summary
The discussion revolves around a physics homework problem involving a beam oscillating in simple harmonic motion (SHM) with a mass that changes when a sack of gravel falls off. Participants clarify the calculations for frequency and amplitude after the sack falls, noting that the original frequency of 0.8 Hz is correct but the amplitude and energy calculations need adjustment. There is confusion regarding the gravitational potential energy (GPE) formula used, with suggestions to correct the reference point for GPE. Additionally, the correct conversion of units from centimeters to meters is emphasized, as the amplitude should be 0.4 m instead of 0.04 m. The conversation highlights the importance of accurate calculations and understanding of SHM principles.
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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


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

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.

k = \omega^2 m

k = 144 N/m

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

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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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.
 
Thanks a lot! haha
 
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?
 
also 40 cm is .4 m not .04 m why would you use .4
 
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.
 
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