A weight hanging from the end of a spring is pulled 8 inches below its rest position. When released, the weight then exhibits simple harmonic motion, with period 1/2 second. Find an equation describing its motion
the amplitude is 8, so
y = 8cos(kx) or 8sin(kx)
It starts with y(0) = -8, so
y = -8cos(kx)
The period is 1/2, so 2π/k = 1/2
y = -8cos(4πx)
To find an equation describing the motion of the weight, we can use the equation for the displacement of an object undergoing simple harmonic motion:
x(t) = A * cos(ωt + φ)
Where:
- x(t) is the displacement of the weight at time t
- A is the amplitude of the motion
- ω is the angular frequency of the motion
- φ is the phase angle
In this case, we are given the period of the motion, which is 1/2 second. The period is related to the angular frequency (ω) through the equation:
T = 2π / ω
Rearranging the equation, we find:
ω = 2π / T
Substituting the given period of 1/2 second, we have:
ω = 2π / (1/2) = 4π rad/s
Next, we need to find the amplitude (A) of the motion. The amplitude is the maximum displacement of the weight from its equilibrium position. In the given information, it is mentioned that the weight is pulled 8 inches (0.67 feet) below its rest position, which means the amplitude is 0.67 feet.
Finally, we need to find the phase angle (φ). The phase angle represents any initial displacement at t = 0. Since the weight is at its maximum displacement when released, the phase angle would be 0.
Plugging these values into the equation for the displacement, we get:
x(t) = 0.67 * cos(4πt)
So, the equation describing the motion of the weight is x(t) = 0.67 * cos(4πt).