1.

A uniform rope of length L and mass m, hangs vertically from a rigid support. A block of mass m_(2)is attached to the free end of the rope. A transverse pulse of wavelengthlamda_(1)is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is lamda_(2). . The ratio lamda_(2)//lamda_(1)is ......

Answer» <html><body><p>`sqrt (( m _(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) + m_(2))/( m _(2)))` <br/>`sqrt ((m _(2))/( m _(1)))` <br/>`sqrt (( m _(1) + m _(2))/( m _(1)))` <br/>`sqrt (( m _(1))/( m _(2)))` </p>Solution :(i) At the lower end of wire, `T_(1) = m _(2) <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> ` <br/> (ii) At the <a href="https://interviewquestions.tuteehub.com/tag/upper-721698" style="font-weight:bold;" target="_blank" title="Click to know more about UPPER">UPPER</a> end of wire, `T_(2) = (m _(1) + m _(2)) g ` <br/> Now, we have `v = f <a href="https://interviewquestions.tuteehub.com/tag/lamda-536483" style="font-weight:bold;" target="_blank" title="Click to know more about LAMDA">LAMDA</a> = sqrt ((T)/(<a href="https://interviewquestions.tuteehub.com/tag/mu-566056" style="font-weight:bold;" target="_blank" title="Click to know more about MU">MU</a>))` <br/> `therefore v _(1) = f lamda_(1) = sqrt ((T_(1))/( mu))` <br/> `v _(2) = f lamda_(2) = sqrt ((T_(2))/( mu ))` <br/> Taking ratio, `(lamda _(1))/( lamda _(2)) = sqrt (( T _(1))/( T _(2)) ) = sqrt (( m _(2) g )/( (m _(1) + m _(2)) g ))` <br/> `therefore (lamda _(2))/( lamda _(1)) = sqrt (( m _(1) + m _(2))/( m _(2)))`<br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P2_C15_E05_019_S01.png" width="80%"/></body></html>


Discussion

No Comment Found

Related InterviewSolutions