What Is the Maximum Speed in This Simple Harmonic Motion Problem?

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Homework Help Overview

The problem involves a mass oscillating between two springs in simple harmonic motion (SHM), with specific parameters including the distance between oscillation points and the frequency of oscillations. Participants are trying to determine the maximum speed of the mass during its motion.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of frequency and maximum speed, with attempts to derive the maximum speed using the amplitude and angular frequency. Some express confusion over their results and question the correctness of their calculations.

Discussion Status

There is an ongoing exploration of the calculations involved in determining the maximum speed. Some participants have provided their calculations, while others have suggested that the answer key may be incorrect. The discussion reflects a mix of attempts and clarifications without reaching a definitive conclusion.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may include specific formatting or presentation requirements for their solutions. There is also an indication of potential discrepancies in the provided answer key.

koat
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A mass between 2 springs moves with shm.
it oscillates between 2 points 5cm apart and completes 40 oscillations in 1min
whats its max speed?

Please help i always get the wrong answer
The answer is 0.44 but why?
 
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koat said:
A mass between 2 springs moves with shm.
it oscillates between 2 points 5cm apart and completes 40 oscillations in 1min
whats its max speed?

Please help i always get the wrong answer
The answer is 0.44 but why?

Show your solution attempt. We can't tell what's going wrong if we don't see what you're doing.
 
2pi*40/60= 4/3 pi
f= 4/3pi*1/2= 2/3
vmax= 2pi*2/3*2.5*10^-2
but the answer is wrong :(
 
koat said:
2pi*40/60= 4/3 pi
f= 4/3pi*1/2= 2/3
vmax= 2pi*2/3*2.5*10^-2
but the answer is wrong :(

Actually, your answer is correct. Looks like the answer key is wrong :smile:

You've correctly calculated the frequency of the oscillations:

## A = 5 cm/2 ##

## f = 40 cycles/min ##

##\omega = \frac{40 cycles}{min} \times \frac{1\;min}{60\;sec} \times \frac{2 \pi \; rad}{cycle} = \frac{4}{3}\pi \frac{rad}{sec}##

## v = A \omega = \frac{10 \pi}{3}\frac{cm}{sec} = 0.105 m/s ##
 

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