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?
To find the furthest distance of the mass from its equilibrium point, we need to find the maximum value of y. In this case, y is given by the formula 2zcos(14πwt), where z, w, and t are positive constants.
The maximum value of the cosine function is 1, which occurs when the argument of the cosine function is 0. Therefore, to find the furthest distance, we need to find the value of t that makes the argument of the cosine function 0.
The argument of the cosine function is 14πwt. To make it 0, we need to solve the equation 14πwt = 0.
Since w and t are positive constants, the value of t that makes the argument 0 is t = 0. Therefore, the furthest distance of the mass from its equilibrium point is 2zcos(0) = 2z.
So, the furthest distance of the mass from its equilibrium point is 2z centimeters.
To find how many oscillations are completed in 1 second, we need to find the frequency of the oscillation.
The frequency, f, is given by the equation f = 1/T, where T is the period of the oscillation.
In this case, the period is the time it takes for one complete oscillation, which is the time it takes for the cosine function to complete one cycle.
The cosine function completes one cycle when its argument increases by 2π. In this case, the argument is 14πwt.
So, the time it takes for the cosine function to complete one cycle is T = (2π)/(14πw) = 1/(7w).
Therefore, the frequency f = 1/T = 7w.
So, in 1 second, the number of oscillations completed is equal to the frequency.
Therefore, the number of oscillations completed in 1 second is 7w.