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kow [346]
3 years ago
14

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) di

splacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?
Physics
1 answer:
gulaghasi [49]3 years ago
6 0

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

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