A particle moving along the x axis in simple harmonic motion starts from its equ

Question

A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 2.90 cm, and the frequency is 2.70 Hz. (f) Determine the total distance traveled between t = 0 and t = 0.56 s.

A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 2.90 cm, and the frequency is 2.70 Hz.

(a) Find an expression for the position of the particle as a function of time. (Use the following as necessary: t, and .)
x =

2.9sin(5.4t)

JUST NEED THE F) PORTION. EVERYTHING ELSE I HAVE CORRECT

Explanation / Answer

x = 2.9sin(2*pi*2.7*t)
time period, T = 1/2.7 = 0.37 s
so 3T/2 = 0.55 s
so, DIstance travelled = 6*A = 6*2.9 = 17.4 cm [ because distance travelled in one T = 4A]

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