A copper rod (length �= 2.0 m, radius =
3.0 *� 10^-�3 m) hangs down from the ceiling. A 9.0-kg object is attached to the lower end of the rod. The rod acts as a “spring,” and the object oscillates vertically with a small amplitude. Ignoring the rod’s mass, find the frequency f of the simple harmonic motion.
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To find the frequency of the simple harmonic motion of the copper rod, we need to make use of the equation for the period of an oscillating system.
The period of simple harmonic motion is given by the formula:
T = 2π√(m/k)
Where:
T is the period,
m is the mass attached to the spring,
k is the spring constant.
Since we want to find the frequency, which is the reciprocal of the period, we can rearrange the formula as:
f = 1/T = 1/(2π√(m/k))
In this case, the mass attached to the rod is 9.0 kg. However, we need to find the spring constant, k, of the copper rod.
The equation for the spring constant is:
k = (4π^2 * m)/L
Where:
m is the mass attached to the spring,
L is the length of the spring.
In this case, the length of the copper rod is given as 2.0 m.
So, substituting the values into the equation, we have:
k = (4π^2 * 9.0 kg)/(2.0 m)
Simplifying the equation, we get:
k = (36π^2 kg)/2.0 m = 18π^2 kg/m
Now we can substitute the value of the spring constant, k, into the formula for frequency:
f = 1/(2π√(m/k))
= 1/(2π√(9.0 kg/(18π^2 kg/m)))
= 1/(2π√(0.5/π^2 m))
= 1/(2√(0.5/π^2 m))
= 1/(2√(2π^2/π^2 m))
= 1/(2√(2 m))
= 1/(2√(2) m)
= 1/(2 * √2) ≈ 0.354 Hz
Therefore, the frequency of the simple harmonic motion is approximately 0.354 Hz.