Question

The displacement of a damped harmonic oscillator is given by

x(t) = e^-0.1t·cos (10πt + ϕ). Here t is in seconds.

The time taken for its amplitude of vibration to drop to half of its initial value is close to:
1. 4 s
2. 7 s
3. 13 s
4. 27 s

Answer 1

Correct Answer - Option 2 : 7 s

Concept:

Amplitude:

For an object in periodic motion, the amplitude is the maximum displacement from equilibrium. The unit for amplitude is meters (m).

Amplitude of oscillation is given by the formula,

A = A₀ e^-0.1t s

Calculation:

Given,

Displacement of the oscillator,

x(t) = e^-0.1t·cos(10πt + ϕ).

Amplitude after t seconds, \(A = \frac{{{A_0}}}{2}\)

Amplitude of oscillation is given by the formula,

A = A₀ e^-0.1t  s

Amplitude of vibration at time t = 0 is given by,

A = A₀ e^-0.1t

A = A₀ e^-0.1×0

A = A₀ × 1 = A₀ [∵ e⁰ = 1]         

Thus, A = A₀ at t = 0

Also, at t = t, if \({\rm{A}} = \frac{{{{\rm{A}}_0}}}{2}\)

A = A₀ e^-0.1t

\(\frac{{{A_0}}}{2} = {A_0}{{\rm{e}}^{ - 0.1t}}\) 

\(\frac{1}{2} = {{\rm{e}}^{ - 0.1{\rm{t}}}}\) 

e^0.1t = 2

0.1t = ln2  

\(t = \frac{{\ln 2}}{{0.1}}\) 

t = 10 ln2

t = 10 ln 2 = 10 × 0.6931

t = 6.931 ≃ 7 s

Therefore, the time taken for its amplitude of vibration to drop to half of its initial value is close to 7 s.