PSE1 13 P026. An object, oscillating with simple harmonic motion, has a period o

Question

PSE1 13 P026. An object, oscillating with simple harmonic motion, has a period of 4.0 s and an amplitude of 0.28 m (a) How long does the object take to move from x = 0.0 m to x-0.06 m? (Assume the oscillator moves directly from 0.0 to 006 m without passing through negative values.) (b) Would the object take the same time to move from x 0.06 m to x 0.12 m? Explain O Yes: the oscillator moves at a constant speed, so it travels equal distances in equal times. O Yes: the oscillator moves at a constant acceleration, so it travels equal distances in equal times. O Yes: the period does not change, so the oscillator travels equal distances in equal times. No; the period does not change, so the oscillator does not travel equal distances in equal times. No the oscillator does not move at a constant speed, so it does not travel equal distances in equal times Need Help lty 4. 3/3 points P t its position varies according to the In an engine, a piston oscilates vith simple harmenic motion so x=5.00 cos ( t + 6 hare x is in centimeters and is in seconds. ind the position of the piston,

Explanation / Answer

given T = 4 s

A = 0.28 m

a. the equation of motion can be written as

x = Asin(wt)

w = 2*pi/T = 1.57 rad/s

hence

x = 0.28sin(1.57t)

so x = 0 at t = 0

x = 0.06 m

0.06 = 0.28*sin(1.57*t)

t = 0.137554 s

b. x = 0.12 m

0.12 = 0.28*sin(1.57*t')

t' = 0.2821 s

hence

t' - t = 0.1445549 s

hence the osscilator takes different time to cover this interval as it has varying speed

hecne option E.

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