Question

An object attached to a spring executes Simple Harmonic Motion (SHM) according to the equation x(t) = [8.8 cos(8t)] mi. Find the:

i. Amplitude
ii. Frequency in Hertz, Hz
iii. Position, x at time t = 88 s
iv. Maximum velocity

Answer 1

The amplitude is 8.8 mi, the frequency is 8 Hz, the position at time t = 88 s is -8.8 mi, and the maximum velocity is 70.4 mi/s.

Step-by-step explanation:

In simple harmonic motion (SHM), the equation for displacement is given by x(t) = 8.8cos(8t) mi. From this equation, we can determine the properties of the motion:

i. The amplitude (A) is the maximum displacement of the object from its equilibrium position. In this case, the amplitude is A = 8.8 mi.

ii. The frequency (f) is the number of complete oscillations per second. It is given by f = 1/T, where T is the period of the motion. In this case, the frequency is f = 8 Hz.

iii. To find the position at time t = 88 s, we substitute t = 88 s into the equation x(t) = 8.8cos(8t) mi. x(88) = 8.8cos(8(88)) = -8.8 mi.

iv. The maximum velocity occurs when the displacement is maximum. The maximum velocity (vmax) can be found using the equation vmax = wA, where w is the angular frequency. In this case, w = 8 rad/s and A = 8.8 mi. Therefore, vmax = 8(8.8) = 70.4 mi/s.