Question

A 0.835 kg block oscillates on the end of a spring whose spring constant is k=41 N/m. The mass moves in a fluid which offers a resistive force F=−bv , where b=0.662 Ns/m.

Part A: What is the period of the motion? (already answered) T= 0.898 s

Answer 1

The period of the oscillatory motion for a 0.835 kg block on a spring with a spring constant
`\(k = 41 \, \text{N/m}\)` and experiencing a resistive force
`\(F = -bv\)`with
`\(b = 0.662 \, \text{Ns/m}\) is \(T = 0.898 \, \text{s}\).`

Step-by-step explanation:

The period
`(\(T\))` of the oscillatory motion of a mass-spring system in a resistive medium can be determined by the formula:

`\[ T = 2\pi \sqrt{(m)/(k - (b^2)/(4m))} \]`

where
`\(m\)` is the mass of the block,
`\(k\)` is the spring constant, and
`\(b\)` is the coefficient of the resistive force.

In this case, substituting the given values
`(\(m = 0.835 \, \text{kg}\), \(k = 41 \, \text{N/m}\), \(b = 0.662 \, \text{Ns/m}\))` into the formula, we obtain:

`\[ T = 2\pi \sqrt{(0.835)/(41 - ((0.662)^2)/(4 * 0.835))} \]`

`\[ T \approx 0.898 \, \text{s} \]`

This result represents the period of the oscillatory motion, indicating the time it takes for the block to complete one full cycle of oscillation. The presence of the resistive force alters the period compared to an ideal mass-spring system without damping.

Understanding the period is crucial in analyzing and designing systems involving harmonic oscillators, as it provides insights into the behavior of the system over time. In this case, the resistive force introduces damping, affecting the rate at which the oscillations occur.