A oscillator undergoes simple harmonic motion along the x-axis with amplitude 3.

Question

A oscillator undergoes simple harmonic motion along the x-axis with amplitude 3.0 m and frequency 1.4 Hz. Which equation could describe the oscillator's motion as a function of time, t?

A. x = (3.0 m) sin (1.4 s-1 t)

B. x = (3.0 m) sin (8.8 s-1 t)

C. x = (6.0 m) sin (1.4 s-1 t)

D. x = (6.0 m) sin (4.4 s-1 t)

An oscillator undergoes simple harmonic motion along the x-axis between positions x = -A and x = +A. At what position is oscillator's kinetic energy 3 times it's elastic potential energy?

A. A/3

B. A/9

C. A/2

D. A/6

Explanation / Answer

As x = A sin (w*t) is the equation of SHM where A= amplitude=3m and w=2*pi*f=2*3.14*1.4=8.79

So B. x = (3.0 m) sin (8.8 s-1 t)

For the said condition, 0.5mv^2= three times of 0.5kx^2

Total energy remains same at all points

So, 3*0.5kx^2 + 0.5kx^2 = 0.5kA^2

Or, 4x^2 = A^2

Or, 2x=A

So (c) x = A/2

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