Question

A mass is oscillating on the end of a spring. The distance, y, of the mass from its equilibrium point is given by the formula y=4zcos(8πwt) where y is in centimeters, t is time in seconds, and z and w are positive constants.

a. What is the furthest distance of the mass from its equilibrium point?
b. How many oscillations are completed in 1 second?

Answer 1

(a.) 4z

(b.) 4w

Step-by-step explanation:

From the equation y=4zcos(8πwt), where z and w are positive constants.

Comparing this equation to the equation of a wave y = Acos(Wt), where A is the amplitude (largest distance from equilibrium) and W is the angular frequency (W=2πf)

(a.) Comparing our wave equation with the given equation, we see that A = 4z in this case (furthest distance of the mass from equilibrium)

(b.) Comparing similarly we can see from our given equation that angular frequency W =8πw we also know that W = 2πf from our wave equation, therefore 2πf = 8πw

Solving for f we have f = 8πw÷2π

f = 4w (Proves our second answer because the frequency is the number of oscillations completed per second)