In a given oscillatory motion the amplitude decreases with time. It is 0.01 m at t = 0 and 0.001 m at t = 100s. Find the logarithmic decrement when the period is 15.61 s. Find the period of oscillations when the damping is zero.
To find the logarithmic decrement, we can use the formula:
Λ = ln(A₁/A₂) / (2πn)
where:
Λ is the logarithmic decrement,
A₁ is the amplitude at time t₁,
A₂ is the amplitude at time t₂,
n is the number of complete cycles between t₁ and t₂.
Given:
A₁ = 0.01 m at t = 0
A₂ = 0.001 m at t = 100s
n = 100s / 15.61s = 6.41 (approximated)
Using the formula, we can calculate the logarithmic decrement:
Λ = ln(0.01/0.001) / (2π * 6.41)
Λ ≈ ln(10) / (2π * 6.41)
Now, let's calculate Λ:
Λ ≈ 0.2301
Therefore, the logarithmic decrement is approximately 0.2301.
To find the period of oscillations when the damping is zero, we can use the equation for the period of an oscillator:
T = 2π * √(m/k)
where:
T is the period,
m is the mass of the oscillating object,
k is the spring constant.
When there is no damping, the oscillation is called simple harmonic motion (SHM). In SHM, the restoring force is directly proportional to the displacement from the equilibrium position (F = -kx). Therefore, the period of SHM is independent of the amplitude.
Since the question does not provide the values of mass or spring constant, we cannot determine the exact period of oscillations when the damping is zero. We need additional information to calculate the period of oscillation in this case.