The position of a mass oscillating on a spring is given by x=(5.8cm)cos[2t/(0.64

Question

The position of a mass oscillating on a spring is given by x=(5.8cm)cos[2t/(0.64s)]

Part A

What is the frequency of this motion?

Express your answer using two significant figures.

Part B

When is the mass first at the position x=5.8cm?

Express your answer using two significant figures.

2) The position of a mass oscillating on a spring is given by x=(3.3cm)cos[2t/(0.80s)].

Part A

What is the period of this motion?

Express your answer using two significant figures.

Part B

What is the first time the mass is at the position x=0?

Express your answer using two significant figures.

3)A ball rolls on a circular track of radius 0.70 m with a constant angular speed of 1.4 rad/s in the counterclockwise direction.

Part A

If the angular position of the ball at t= 0 is = 0, find the x component of the ball's position at the time 2.6 s . Let = 0 correspond to the positive x direction.

Express your answer using two significant figures.

Part B

Find the x component of the ball's position at the time 5.1 s .

Express your answer using two significant figures.

Part C

Find the x component of the ball's position at the time 7.5 s

Express your answer using two significant figures.

Thanks for the help!

Explanation / Answer

1.)

Part -A

x=(5.8cm)cos[2t/(0.64s)]

cos(0) = 1 , and is only 1 again (meaning it has undergone a full period) at 2.
So, 2 *pi* t/0.64= 2*pi when t = .64 sec. Therefore the period is 0.64 sec

frequency = 1/T { T is time period }

= 1/0.64 =1.5625 s^-1

Part -B

x=(5.8cm)cos[2t/(0.64s)]

Next, cos() = cos(3) = -1. So the first time this happens is at cos().
So we need to know when 2 t/0.64 =
Solving, we get t = 0.64/2 = 0.32 sec .

2.)

Part -A

cos(0) = 1 , and is only 1 again (meaning it has undergone a full period) at 2.
So, 2 ** t/0.8= 2* when t = .80 sec. Therefore the period is 0.80 sec

Part-B

Next, cos(/2) = cos(3/2) = 0. So the first time this happens is at cos(/2).
So we need to know when 2 t/0.8 = /2
Solving, we get t = 0.8/4 = 0.2 sec .

3.)

at the times 2.6 s,
rad = 1.4 * 2.6 = 3.64 rad = 208.556 deg     { rad = 180 deg }
x = 0.70 * cos 208.56 = - 0.6148 m

at the times 5.1 s,
rad = 1.4 * 5.1 = 7.14 rad = 409.09 deg
x = 0.70 * cos 409.09 = 0.458m

at the times 7.5 s,
rad = 1.4 * 7.5 = 10.5 rad = 601.6 deg
x = 0.70 * cos 601.6 = - 0.332 m

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