The angular frequency of the damped oscillator is given by $\omega = \sqrt{\frac{k}{m} - \frac{r^2}{4m^2}}$,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:

  • A
    increases by $1\%$
  • B
    increases by $8\%$
  • C
    decreases by $1\%$
  • D
    decreases by $8\%$

Explore More

Similar Questions

In damped oscillations,the damping force is directly proportional to the speed of the oscillator. If the amplitude becomes half of its initial value $A_0$ in $1 \, s$,then after $2 \, s$,the amplitude will be:

Which of the following figures represents damped harmonic motion?

$A$ block of mass $1 \, kg$ attached to a spring is made to oscillate with an initial amplitude of $12 \, cm$. After $2 \, minutes$ the amplitude decreases to $6 \, cm$. Determine the value of the damping constant $b$ for this motion. (Take $\ln 2 = 0.693$)

The amplitude of a damped oscillator becomes half in $1$ minute. The amplitude after $3$ minutes will be $\frac{1}{x}$ times the original. Then $x$ is

The amplitudes of a damped harmonic oscillator after $2 \ s$ and $4 \ s$ are $A_1$ and $A_2$ respectively. If the initial amplitude of the oscillator is $A_0$,then

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo