1.

Two pendulums of lengths 100 cm and 121 cm suspended side by side. A suspended bobs suspended at its end stretched and leave together. After how many minimum oscillations of big pendulum, both are in same phase?

Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a>`<br/>`10`<br/>`21`<br/>`20`</p>Solution :Suppose `T_(1)" and "T_(2)` are the <a href="https://interviewquestions.tuteehub.com/tag/periodic-598580" style="font-weight:bold;" target="_blank" title="Click to know more about PERIODIC">PERIODIC</a> time of two pendulums <br/> `T_(1)= 2pi sqrt((100)/(g))" and "T_(2)= 2pi sqrt((121)/(g))""[therefore T_(1) <a href="https://interviewquestions.tuteehub.com/tag/lt-537906" style="font-weight:bold;" target="_blank" title="Click to know more about LT">LT</a> T_(2)," because "l_(1) lt l_(2)]` <br/> Suppose at t=0, both oscillate at same time but their oscillations are the same due to different period. <br/> Suppose pendulum of long string <a href="https://interviewquestions.tuteehub.com/tag/makes-2811811" style="font-weight:bold;" target="_blank" title="Click to know more about MAKES">MAKES</a> n oscillations and pendulum of short string makes `(n+1)` oscillations. <br/> `therefore (n+1)T_(1) = nT_(2)` <br/> `therefore (n+1)xx 2pi sqrt((100)/(g))= n xx 2pi sqrt((121)/(g))` <br/> `therefore (n+1)10= 11n` <br/> `therefore 10n +10 = 11n` <br/> `therefore n = 10`.</body></html>


Discussion

No Comment Found

Related InterviewSolutions