JEE Main 2014 · 11 Apr · Shift 1 · Physics Q14

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PhysicsSingle correct+4 / -1Question 14 of 90

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

increases by 1%

increases by 8%

decreases by 1%

decreases by 8%

Answer and worked solution

Official answer: A

$\omega=\sqrt{\frac km}\sqrt{1-\frac{r^2}{4mk}}=\omega_0\sqrt{1-0.02}\approx\omega_0(1-0.01).$
$\omega$ falls by 1%, so the period increases by 1%.

Topics:Damped and forced oscillations

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JEE Main 2014 · 11 Apr · Shift 1, Physics question 14