Question

A body vibrating with simple harmonic motion has a frequency of 4Hz and amplitude of 0.15m. Calculate the maximum values of the acceleration and velocity.
Calculate the acceleration and velocity at a point 0.09m from the equilibrium position of motion

Answer 1

Solution. We write the equation simple harmonic motion

$$x(t)=Asin(2\pi ft)$$

where x(t) is displacement; A=0.15m is amplitude; t is time; f=4Hz is frequency.

We write the equation of the dependence of velocity on time

$$v(t)=\frac{dx}{dt}=2\pi fAcos(2\pi ft)$$

Therefore maximum values of the velocity

$$v_{max}=2\pi fA=2\pi\times 4\times 0.15=3.77 \frac{m}{s}$$

We write the equation of the dependence of acceleration on time

$$a(t)=\frac{dv}{dt}=-4\pi^2 f^2Asin(2\pi ft)$$

Therefore maximum values of the velocity

$$a_{max}=4\pi^2f^2A=4\pi^2 \times 4^2\times0.15=30.16 \frac{m}{s^2}$$

Find a time that corresponds to a displacement of 0.09 m

$$0.09=0.15sin(8\pi t) \implies sin(8\pi t)=0.6 \implies t=0.0256s$$

At this point in time, velocity and acceleration are respectively equal

$$v=3.77cos(2\pi \times 4 \times 0.0256)=3.016\frac{m}{s}$$

$$a=-30.16sin(2\pi\times4 \times 0.256)=-18.096 \frac{m}{s^2}$$

Answer.

$$v_{max}=3.77 \frac{m}{s}$$

$$a_{max}=30.16 \frac{m}{s^2}$$

The acceleration and velocity at a point 0.09m from the equilibrium position of motion is equal to

$$v=3.016\frac{m}{s}$$

$$a=-18.096 \frac{m}{s^2}$$