A copper rod (length = 1.3 m, radius = 3.70 10-3 m) hangs down from the ceiling. A 8.3-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.

Answer in Hz

To find the frequency of the simple harmonic motion, we can use the formula:

f = (1/2π) * √(k/m)

where:
- f is the frequency
- π is a mathematical constant (approximately equal to 3.14159)
- k is the spring constant of the rod
- m is the mass of the object attached to the rod

The spring constant (k) can be calculated using the formula:

k = (G * A) / L

where:
- G is the shear modulus of the material (for copper, G is approximately 4 × 10^10 Pa)
- A is the cross-sectional area of the rod
- L is the length of the rod

To calculate A, we can use the formula for the area of a circle:

A = π * r^2

where:
- r is the radius of the rod

Given:
- Length of the rod (L) = 1.3 m
- Radius of the rod (r) = 3.70 × 10^(-3) m
- Mass of the object (m) = 8.3 kg
- Shear modulus of copper (G) = 4 × 10^10 Pa

Let's calculate the frequency step-by-step:

1. Calculate the cross-sectional area, A:

A = π * r^2
= π * (3.70 × 10^(-3))^2
≈ 4.29 × 10^(-5) m^2

2. Calculate the spring constant, k:

k = (G * A) / L
= (4 × 10^10 Pa) * (4.29 × 10^(-5) m^2) / (1.3 m)
≈ 1.305 × 10^6 N/m

3. Calculate the frequency, f:

f = (1/2π) * √(k/m)
= (1/(2π)) * √((1.305 × 10^6 N/m) / (8.3 kg))
≈ 8.54 Hz

Therefore, the frequency (f) of the simple harmonic motion is approximately 8.54 Hz.

To find the frequency of the simple harmonic motion, we need to use the formula for the oscillation period of a mass-spring system:

T = 2π√(m/k)

where T is the period, m is the mass attached to the spring, and k is the spring constant.

In this case, the copper rod acts as a spring, so we need to find its effective spring constant.

The spring constant of a rod can be calculated using the formula:

k = (pi * E * r^4) / (4 * L^3)

where E is the Young's modulus of the material (for copper, approximately 1.2 * 10^11 N/m^2), r is the radius of the rod, and L is the length of the rod.

Let's plug in the given values:

E = 1.2 * 10^11 N/m^2
r = 3.70 * 10^-3 m
L = 1.3 m

k = (pi * (1.2 * 10^11 N/m^2) * (3.70 * 10^-3 m)^4) / (4 * (1.3 m)^3)

Now we can find the spring constant.

Next, we need to find the effective mass for the system, which includes the mass of the object (8.3 kg) and the mass of the rod (which we are ignoring).

The total mass of the system is:

m_total = m_object + m_rod

Since we are ignoring the mass of the rod, the effective mass becomes:

m_effective = m_object

Now, we can plug the values into the equation for the period:

T = 2π√(m_effective / k)

T = 2π√(8.3 kg / k)

Finally, the frequency is the inverse of the period:

f = 1 / T

By calculating the spring constant and plugging in the values, we can find the frequency f of the simple harmonic motion in Hz.