Question

A metal block with a mass of 4.00 kg oscillates at the end of a vertical spring with a spring constant of 2.20 ✕ 10⁴N/m. The motion is damped by air resistance, and the damping coefficient is b = 3.00 N · s/m.

What is the frequency (in Hz) of the damped oscillation?

Answer 1

The frequency of the damped oscillation for a 4.00 kg metal block attached to a vertical spring with a spring constant of 2.20 × 10⁴ N/m and damped by air resistance with a damping coefficient of b = 3.00 N · s/m is approximately 11.80 Hz.

Step-by-step explanation:

To calculate the frequency of the damped oscillation for a metal block with a mass of 4.00 kg, attached to a vertical spring with a spring constant of 2.20 × 10⁴ N/m, and damped by air resistance with a damping coefficient b = 3.00 N · s/m, we will use the formula for the frequency of a damped harmonic oscillator. The damped frequency (f_d) is given by:

f_d = (1/2π) √((k/m) - (b²/(4m²)))

First, we identify the undamped angular frequency:

ω_0 = √(k/m)

For the given values k = 2.20 × 10⁴ N/m and m = 4.00 kg:

ω_0 = √(2.20 × 10⁴ N/m / 4.00 kg) = √(5500 s⁻²) = 74.16 rad/s

Now, we calculate the damped angular frequency (ω_d):

ω_d = √(ω_0² - (b²/(4m²)))

ω_d = √((74.16 rad/s)² - (3.00 N·s/m)² / (4 × (4.00 kg)²))

ω_d = √(5500 - 0.140625) rad/s

ω_d ≈ 74.16 rad/s (since 0.140625 is very small compared to 5500, we can ignore it for a rough approximation)

Finally, we convert to frequency in Hz:

f_d = ω_d / (2π) ≈ 74.16 rad/s / (2π) ≈ 11.80 Hz

The frequency of the damped oscillation is approximately 11.80 Hz.