A passenger car traveling down a rough road bounces up and down at 1.3 Hz with a

Question

A passenger car traveling down a rough road bounces up and down at 1.3 Hz with a maxium vertical acceleration of 0.2 m/s^2. both typical values.

A. What is the amplitude?

b. What is the maximum speed of oscilation?

A spring has an unstretch-length of 12 cm. When an 80g ball is hung from it, the length increses by 4 cm, the ball is then pulled down another 4 cm and then it is released.

a. What is the spring constant?

b. What is the period of the oscillation?

Please state what the known values are and the equations used. Exaplin EVERYTHING in step by step form for me to understand. Thanks.

Explanation / Answer

1. f = 1.3 Hz
a_max = 0.2 m/s^2

a_max = ?^2Y
0.2 = (2*pi*f)^2*Y
Y = 0.2 / 66.77
Y = 0.0029954 m

V_max = ?Y = 2*pi*f*Y
V_max = 2*pi*1.3*0.0029954
V_max = 0.02447 m/s

2. The equation for the force of a spring is: F = -k*(dx)

k - spring constant
dx - change in elongation of spring

The force of gravity on the ball has to be the same as the force on the spring (when at rest), so

mg = k*(dx) (you can drop the negative sign since you know the forces are in opposite directions)

When you plug in all the numbers, you get k = 19.6 N/m

(pay attention to units... the easiest thing to do is change everything to meters and kilograms)

For the second part, the frequency of the motion is:

T = 2*pi*(m/k)^0.5

You now have both m (mass) and k, so you can plug in and get

T = 0.401 s

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