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=6zcos(10wt)
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?
distance = cm
(b) How many oscillations are completed in 1 second?
number of oscillations =
(a) 6z is the amplitude
(b) 2 pi f t = 10 w t
f = 5/(2 pi) is the frequency, which is what they are asking for
(a) The furthest distance of the mass from its equilibrium point can be determined by finding the amplitude of oscillation, which is represented by the constant z in the formula. However, since there is no specific value provided for z in the question, I'm afraid I can't calculate the exact distance in centimeters. Maybe the spring has some commitment issues and doesn't want to reveal its true amplitude!
(b) To determine the number of oscillations completed in 1 second, we need to consider the angular frequency represented by the constant w in the formula. However, without the specific value of w, I can't provide you with a precise number of oscillations. It seems like our spring is keeping its cycle count a secret, maybe it's worried about its personal growth!
(a) 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 equation y = 6zcos(10wt).
The maximum value of the cosine function is 1.
Therefore, the furthest distance of the mass from its equilibrium point is y = 6z(1) = 6z cm.
(b) The formula for the number of oscillations completed in one second is given by 1 / T, where T is the period of oscillation. The period is the time taken to complete one full oscillation.
In this case, the angular frequency is 10w, where w is a positive constant.
The period (T) is given by T = 2π / (10w).
Therefore, the number of oscillations completed in one second can be calculated as:
number of oscillations = 1 / (2π / (10w)) = 10w / 2π.
So, the number of oscillations completed in one second is 10w / 2π.
To answer these questions, let's analyze the given formula for the distance of the mass from its equilibrium point: y = 6zcos(10wt).
(a) The furthest distance of the mass from its equilibrium point can be found by considering the amplitude of the oscillation. In this case, the amplitude is given by z. Therefore, the furthest distance is equal to the absolute value of the amplitude (z). So, the answer is given by:
distance = |z| cm.
(b) To determine the number of oscillations completed in 1 second, we need to find the frequency of the oscillation. From the given formula, we can see that the frequency of oscillation is 10w oscillations per second. Therefore, the number of oscillations completed in 1 second is equal to 10w. So, the answer is:
number of oscillations = 10w.