The angular frequency of the damped oscillator is given by, ω = √((k/m-(r²)/(4m²))) where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio r²/mk is 8%, the change in time period compared to the undamped oscillator is approximately as follows: (a) increases by 1% (b) increases by 8% (c) decreases 1% (d) decreases 8%

Answer: (a). ω = √(k/m)√(1 - (r²)/(4mk)) = ω₀√(1 - 0.08/4) = ω₀√(1 - 0.02). ω ≈ ω₀(1 - 0.01), so ω decreases by 1%. T ∝ 1/ω, so T increases by about 1%. So the answer is (a) increases by 1%.

The angular frequency of the damped oscillator is given by, ω =… | JEE Main 2014 (11 Apr, Shift 1)

Q19JEE Main 2014 · 11 APR · SHIFT 1Difficult

Damped, Forced Oscillations & Superposition

The angular frequency of the damped oscillator is given by,

$\omega = \sqrt{\left(\frac{k}{m}-\frac{r^2}{4m^2}\right)}$ where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio $\frac{r^2}{mk}$ is 8%, the change in time period compared to the undamped oscillator is approximately as follows:

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Answer: (a)

$\omega = \sqrt{\frac{k}{m}}\sqrt{1 - \frac{r^2}{4mk}} = \omega_0\sqrt{1 - \frac{0.08}{4}} = \omega_0\sqrt{1 - 0.02}$ .
$\omega \approx \omega_0(1 - 0.01)$ , so $\omega$ decreases by 1%.
$T \propto 1/\omega$ , so $T$ increases by about 1%.
So the answer is (a) increases by 1%.

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