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 2zcos(14πwt) where y is in centimeters, t is time in seconds, w and t and are positive constants.
What is the furthest distance (cm) of the mass from its equilibrium point?
How many oscillations are completed in 1 second?
Hey Stephanie.. if I'm understanding correctly there's a typo, and you actually meant 'w and z are positive constants'?
a) Hint: Look at a picture of a cosine graph and see what value of '14*pi*w*t' maximizes the value
b) Hint: When is 2zcos(14piwt) equal to zero?
If this doesn't help show me your work so far and I'll lend a hand!
To find the furthest distance of the mass from its equilibrium point, we need to determine the maximum value of the formula 2zcos(14πwt). The maximum value of the cosine function is 1, so the furthest distance can be found by multiplying 2z by 1. Therefore, the furthest distance of the mass from its equilibrium point is 2z centimeters.
To find the number of oscillations completed in 1 second, we need to determine the frequency of the oscillation. The frequency is given by the value of w in the formula. Since w is a positive constant, it represents the number of complete oscillations per second.
Therefore, the number of oscillations completed in 1 second is equal to the value of w.