Question

6. + -14 points SerPSE 10 15.A.P.033. An object attached to a spring vibrates with simple harmonic motion as described by the following figure. * (cm) 2.00 1.00 H 0.00 1 2 3 4 5 6 -1.00 -2.00 (a) For this motion, find the amplitude. (b) For this motion, find the period. (c) For this motion, find the angular frequency. rad/s (d) For this motion, find the maximum speed. cm/s (e) For this motion, find the maximum acceleration cm/s2 (f) For this motion, find an equation for its position x as a function of time. (Use the following as necessary: t.) X(t) = cm

Answer 1

a) From the graph we can see the amplitude for this motion is the maximum value of x.

So, for this motion, the amplitude is 2 cm.

b) For this motion, the period is the time to complete a cycle.

So, the period is 4 seconds.

c) Let the angular frequency is $\omega$.

  $\therefore \: \: \: \omega=\frac{2\pi }{T}$

  

or w=" rad/s

iw=1.571 rad/s

$\therefore$ For this motion the angular frequency is 1.571 rad/s.

d) Let the maximum speed is v.

We know maximum speed

V = Aw

or v= 2 x 1.571 cm/s

i v= 3.142 cm/s

$\therefore$ The maximum speed for this motion is 3.142 cm/s

e) Let the maximum acceleration for this motion is a.

We know maximum acceleration

7 =D

or a= (1.571)2 x 2 cm/s

  

1. a = 4.935 cm/s

$\therefore$ The maximum acceleration for this motion is 4.935 cm/s².

f) The equation for its position x as a function of time is

(t) = Asin(wt + o)

Here $\phi$ is phase difference and equals to zero.

  

(t) = 2 sin (1.571t)