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

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SUMMARY

The maximum speed of a mass oscillating in simple harmonic motion (SHM) between two springs, completing 40 oscillations in one minute and oscillating between two points 5 cm apart, is calculated to be 0.105 m/s. The frequency of the oscillations is determined to be 4/3 π rad/sec, and the amplitude is 2.5 cm. The confusion arose from an incorrect answer key, but the calculations provided by the participants confirm the correct maximum speed.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Knowledge of frequency and angular frequency calculations
  • Familiarity with amplitude in oscillatory motion
  • Ability to convert units (cycles per minute to radians per second)
NEXT STEPS
  • Study the principles of simple harmonic motion in detail
  • Learn how to derive maximum speed in SHM using the formula v = Aω
  • Explore the relationship between frequency, period, and angular frequency
  • Practice converting between different units of measurement in physics problems
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking to clarify concepts related to simple harmonic motion.

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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