Question

A cart at the end of a spring undergoes simple harmonic motion of amplitude A = 10 cm and frequency 5.0 Hz. Assume that the cart is at x=−A when t=0.

a) Determine the period of vibration.
b) Write an expression for the cart's position as a function of time.
c) Determine the position of the cart at 0.050 s .
d) Determine the position of the cart at 0.100 s .

Answer 1

a) The period of vibration is 0.2 seconds. b) The expression for the cart's position as a function of time is x(t) = A * sin(2πft). c) The position of the cart at 0.050 seconds is 10 cm. d) The position of the cart at 0.100 seconds is 0 cm.

Step-by-step explanation:

a) The period of vibration can be determined using the formula T = 1/f, where T is the period and f is the frequency. In this case, the frequency is given as 5.0 Hz. So, the period would be T = 1/5.0 Hz = 0.2 seconds.

b) The expression for the cart's position as a function of time can be written as x(t) = A * sin(2πft), where x(t) is the position at time t, A is the amplitude, f is the frequency, and t is the time.

c) To determine the position of the cart at 0.050 seconds, substitute t = 0.050 seconds into the expression for x(t) from part (b). x(0.050) = 10 cm * sin(2π * 5.0 Hz * 0.050 s) = 10 cm * sin(0.5π) = 10 cm * 1 = 10 cm.

d) To determine the position of the cart at 0.100 seconds, substitute t = 0.100 seconds into the expression for x(t) from part (b). x(0.100) = 10 cm * sin(2π * 5.0 Hz * 0.100 s) = 10 cm * sin(π) = 10 cm * 0 = 0 cm.