Question

The function x = (5.2 m) cos[(5πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 5.3 s, what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?

Answer 1

(a) Displacement = - 3.0576 m

(b) Velocity
`=-66.48` m/s

(c)Acceleration = -753.39 m²/s

(d)The phase motion is 26.7
`\pi`.

(e)Frequency =2.5 Hz.

(f)Time period =0.4 s

Step-by-step explanation:

Given function is

`x= (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5]`

(a)

The displacement includes the parameter t, so,at time t=5.3 s

`x|_(t=5.3)= (5.2 m)cos[ (5\pi \ rad/s)5.3+ \frac\pi5]`

`= (5.2 m)cos[ 26.5\pi+ \frac\pi5]`

=(5.2)(-0.588)m

= - 3.0576 m

(b)

`x= (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5]`

To find the velocity of simple harmonic motion, we need to find out the first order derivative of the function.

`v=(dx)/(dt)`

`=(d)/(dt) (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5]`

`= (5.2 m)(-5\pi)sin[ (5\pi \ rad/s)t+ \frac\pi5]`

`= -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5]`

Now we can plug our value t=5.3 into the above equation

`v= -26\pi sin[ (5\pi \ rad/s)5.3\ s+ \frac\pi5]`

`=-66.48` m/s

(c)

To find the acceleration of simple harmonic motion, we need to find out the second order derivative of the function.

`v= -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5]`

`a=(d^2x)/(dt^2)`

`=(dv)/(dt)`

`=(d)/(dt)( -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5])`

`= -26\pi (5\pi)cos[ (5\pi \ rad/s)t+ \frac\pi5]`

`= -130\pi^2cos[ (5\pi \ rad/s)t+ \frac\pi5]`

Now we can plug our value t=5.3 into the above equation

`a= -130\pi^2cos[ (5\pi \ rad/s)5.3 \ s+ \frac\pi5]`

= -753.39 m²/s

(d)

The general equation of SHM is

`x=x_mcos(\omega t+\phi)`

`x_m` is amplitude of the displacement,
`(\omega t+\phi)` is phase of motion,
`\phi` is phase constant.

So,

`(\omega t+\phi)=5\pi t+\frac\pi5`

Now plugging t=5.3s

`(\omega t+\phi)=5\pi * 5.3+\frac\pi5`

=26.7
`\pi`

The phase motion is 26.7
`\pi`.

The angular frequency
`\omega = 5\pi`

(e)

The relation between angular frequency and frequency is

`\omega =2\pi f`

`\therefore f=(\omega)/(2\pi)`

`=(5\pi)/(2\pi)`

`=\frac52`

= 2.5 Hz

Frequency =2.5 Hz.

(f)

The relation between frequency and time period is

`T=\frac1 f`

`=\frac1{2.5}`

=0.4 s

Time period =0.4 s