a particle moves back and forth along the x axis between the points x=0.20m and

Question

a particle moves back and forth along the x axis between the points x=0.20m and x=-0.20m. the period of motion is 1.2s, and it is simple harmonic. At the time t=0, the particle is at x=0 and its velocity is positive.

A) What is the frequency of the motion? The angular frequence?

b) what is the amplitude of motion?

C) what is the phase constant?

D) At what time will the particle reach the point x=0.20m? At what time will it reach the point x= -0.10m?

E) What is the speed of the particle when it is at x=0? What is the speed of the particle when it reaches the point x= - 0.10 m?

For simple harmonic motion, x = A cos (omega l delta) and d2x/dl2 = - omega 2 A cos (omega l + delta). Hence, the peak acceleration is omega 2 A, which yields omega 2 A = 0.4 g or A = 0.4g/ omega 2 = 0.4 Times 9.8 m/s2/(2 pi Times 0.3/s)2 = 1.1 m Thus, the amplitude of the motion is 1.1 m, i.e., 2.2 m from one extreme of the motion to the other.

Explanation / Answer

Part - A

f = 1/T = 1/1.2 = 0.833 Hz

angular = w = 2 x pi x f = 5.233 rad/s

Part - B

amplitude = 0.2 m

Part - C

phase constant = w x t = 5.233 x 1.2 = 6.28 rad

Part - D

You know teh paritcle's position has to be described by:

x(t) = A*sin(2*pi*t/T) + B*cos(2*pi*t/T) where T = period and A,B are constants. You also know that

x(t=0) = x(0) = 0.2m = B since sin(0) = 0 and cos(0) = 1

and

v(0) = speed = 0 =dx/dt|t=0 = 2*pi/T*(A*cos(2*pi*t/T) - B*sin(2*pi*t/T))|t=0

0 = 2*pi/T *A ---> A = 0

So x = 0.2 *cos(2*pi*t/T) meters

Now it reaches x = 0.2m at t = 0, T, 2T, 3T, ...

Ite reaches x = -0.2 m at t= T/2, 3T/2, 5T/2, ....

Now for x = - 0.1m here's what you do. Let x = x1 = -0.1 m then

x1 = 0.2*cos(2*pi*t/T) --> divide by 0.2 and take inverse cosine:

2*pi*t/T = arccos(x1/0.2) ---> simplify

t = T/(2*pi)*arccos(x1/.02)

Now you have to be careful at this point because x1/0.2 = -0.5. This means the angle is a second quadrant angle therefore > 90 degrees so make sure you get 2.0944 radians when you take the inverse cosine.

The you should have:

t = 1.2/(2*pi)*2.0944 sec = 0.4 sec

Part - E

TO find speed at x = 0, first find time t0 by using

0 = 0.2*cos(2*pi*t0/T) now 2*pi*t0/T = pi/2 for cosine to go to zero

t0 = T/4 = 0.3 sec

Then us v(t0) = 2*pi/T* 0.2 sin(2*pi*t0/T) = 0.4*pi/T *sin(2*pi*T/(4T)) = 0.4*pi/T *sin(pi/2) = 0.4*pi/T

v(t0) = 1.05 m/s

For x = -0.1 m, you already know t = 0.4 sec so

v(0.4) = 2*pi/T* 0.2 sin(2*pi*0.4/T) = 0.907 m/s

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