- Parent Category: 02 Newtonian Mechanics
- Category: 09 Oscillations
- Created: Thursday, 20 August 2015 15:00
- Last Updated: Thursday, 25 May 2017 15:47
- Published: Monday, 11 April 2016 15:00
- Hits: 3773
A body moves in a simple harmonic motion and the following graph gives the variation of its displacement x with time t.
a) Write an equation to represent the given simple harmonic motion.
b) Find the time duration in the first cycle when the body is located above a displacement of 0.25 m.
[ x = xo sin (1.05 t) , Δt = 1.9 s ]
The hint can be found in the model
a) x0 = 0.5 m
ω = 1.05
x = x0sin(ω t) = 0.5 sin(1.05t)
b) At x = 0.25 m,
substituting back into the equation,
0.25 = 0.5 sin(1.05t)
solving for t gives
0.5 = sin(1.05t)
solving of the ranging of time, t
1.05t = 0.524 and π - 0.524
t1 = 0.5 , t2 = 2.4 s
looking at the graphical form of the solution of the 2 times, t1 and t2.
giving the range to be t2 - t1 = 2.4 - 1.5 = 1.9 s
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