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

Answer:

(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

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.

$v=\frac{dx}{dt}$

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

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

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

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.

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

$a=\frac{d^2x}{dt^2}$

 $=\frac{dv}{dt}$

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

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

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

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

$(\omega t+\phi)=5\pi \times 5.3+\frac\pi5$

             =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

$\omega =2\pi f$

$\therefore f=\frac{\omega}{2\pi}$

     $=\frac{5\pi}{2\pi}$

    $=\frac52$

   = 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