JEE Mains · Physics · STD 11 - 13. oscillations
The angular frequency of the damped oscillator is given by, \(\omega = \sqrt {\left( {\frac{k}{m} - \frac{{{r^2}}}{{4{m^2}}}} \right)} \) 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
- A increases by \(1\%\)
- B increases by \(8\%\)
- C decreases by \(1\%\)
- D decreases by \(8\%\)
Answer & Solution
Correct Answer
(A) increases by \(1\%\)
Step-by-step Solution
Detailed explanation
The change in time period compared to the undamped oscillator increases by \(1\%\)
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