Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Confused! kindly explain, - Oscillations and Waves - JEE Main-5

 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 :

  • Option 1)

      increases by 1%

  • Option 2)

     increases by 8%

  • Option 3)

      decreases by 1%

  • Option 4)

      decreases by 8%

 

Answers (1)

As we learned

Resultant equation in damped motion -

x=A_{0}.e^{-\frac{bt}{2m}}.\sin \left ( \omega' t+\delta \right )
 

- wherein

\omega'= \sqrt{\frac{K}{m}-\left ( \frac{b}{2m} \right )^{2}}

= \sqrt{\omega {_{0}}^{2}-\left ( \frac{b}{2m} \right )^{2}}

 

 

\omega = \sqrt{\frac{k}{m }-\frac{r^{2}}{4m^{2}}}= \sqrt{\frac{k}{m}\left ( 1-\frac{r^{2}}{4km} \right )}

\omega = \sqrt{\frac{k}{m}}\left ( 1-\frac{r^{2}}{4km} \right )^{1/2}

\omega = \omega _{0}\left [ 1-\frac{r^{2}}{8km} \right ]

        \omega = \omega _{0}\left ( 1-0.01 \right )

    \omega =0.99 \: \: \omega _{0}

T= 1.01\: \: T_{0}

Time period is increases by 1 %


Option 1)

  increases by 1%

Option 2)

 increases by 8%

Option 3)

  decreases by 1%

Option 4)

  decreases by 8%

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’
anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year
anam-khan

What after JEE Main results 2024? All you need to know about JEE Adv, CSAB, JoSAA counselling

Latest Question