Simple Harmonic Motion MCQ Question

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SUMMARY

The discussion revolves around calculating the period of oscillation for a simple pendulum based on its motion between specific marks on a scale. The extreme positions of the pendulum bob are at 300mm and 500mm, with the bob moving from 350mm to 450mm in 1.0s. The relevant equations include T=2π/ω for period calculation and x = x0 sin(ωt) for displacement. The correct approach simplifies the problem by focusing on the amplitude and equilibrium position, leading to a definitive period calculation.

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  • Understanding of simple harmonic motion principles
  • Familiarity with the equations of motion for oscillating systems
  • Knowledge of angular velocity and its relation to period
  • Ability to interpret pendulum motion in terms of amplitude and equilibrium
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  • Study the derivation of the period formula T=2π/ω in detail
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Homework Statement



A simple pendulum was moving in front of a horizontal scale. the extreme positions of the bob were at the 300mm and 500mm marks. The bob moved from the 350mm to the 450mm marks in 1.0s. What is the period of the oscillations?
A. 2.0s
B. 3.0s
C. 4.0s
D. 6.0s

Homework Equations



x = x0 sin(wt + fai) , where "fai" is the phase difference.
fai/2pi = x/wavelength, x is the distance apart between 2 points out of phase.
T=2pi/w , where T is the period and w is the angular velocity.

The Attempt at a Solution



i tried solving by using x0=100,
taking 400 to be the equlibrium point, since it was middle of 300 and 500.
thus, amplitude is x0=100.
so by the formula x = x0 sin(wt + fai),
i still need to find fai and x.
so i used fai/2pi = x/wavelength.
and then I'm stuck.. ):

any help would be appreciated. thank you!
 
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Your starting equation is OK, but the phase is an irrelevant complication here. If you clock starts when the pendulum is at the equilibrium position, the equation is simply

x(t) = A sin(ωt) with A = 100 m.

Note that 350 mm is -A/2 and 450 mm is +A/2. This should help.
 

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