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Show that the velocity of travelling wave produced in a string is v= sqrt((T)/(mu))

Answer» <html><body><p></p>Solution :Velocity of transverse waves in a stretched string: Let us compute the velocity of transverse travelling waves on a string . When a jerk is given at <a href="https://interviewquestions.tuteehub.com/tag/one-585732" style="font-weight:bold;" target="_blank" title="Click to know more about ONE">ONE</a> end (left end ) of the rope , the wave pulses move towards right end with a velocity v . This means that the pulses move with a velocity v and respect to an observer who is at rest frame . Suppose an observer also moves with same velocity v in the direction of motion of the wave pulse , then that observer will notice that the wave pulse is stationary and the rope is moving with pulse with the same velocity v . Consider an elemental segment in the string . Let A and B be two points on the string at an instant of time . Let dl and dm be the length and mass of the elemental string , respectively . By definition , linear mass density `mu` is <br/> `mu = (dm)/(dl) "" ... (1)` <br/> `dm = mu dl "" ... (2)` <br/> The elemental string AB has a curvature which looks like an arc of a circle with centre at 0 , radius Rand the arc subtending an angle `theta`at the origin O. The angle `theta`can be written in termsof arc length and radius as `theta= (dl)/(R)` . The centripetal acceleration supplied by the tension in thestring is <br/> `a_(cp) = (v^(2))/(R) "" ... (3)` <br/> Then , centripetal force can be obtained when mass of the string (dm) is included in equation (3) . <br/> `F_(cp) =((dm) v^(2))/(R) "" ... (4)` <br/> The centripetal force experienced by elemental string can be calculated by substituting equation (2) in equation ( 4) we get <br/>`((dm) v^(2))/(R) = (mu v^(2) dl)/(R) "" ... (5) ` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_V02_C11_E01_033_S01.png" width="80%"/> <br/> The tension T acts along the tangent of the elemental segment of the string at A and B. Since the arc length is very small, variation in the tension force can be ignored. We can resolve T into horizontal component T cos `((theta)/(2))`and vertical component T sin `((theta)/(2))` The horizontal components at A and Bare equal in magnitude but <a href="https://interviewquestions.tuteehub.com/tag/opposite-1137237" style="font-weight:bold;" target="_blank" title="Click to know more about OPPOSITE">OPPOSITE</a> in direction, therefore, they cancel each other. Since the elemental arc length AB is taken to be very small, the vertical components at A and B appears to acts vertical towards the centre of the arc and hence, they add up. The net radial force F is <br/> `F_(r)=<a href="https://interviewquestions.tuteehub.com/tag/2t-301164" style="font-weight:bold;" target="_blank" title="Click to know more about 2T">2T</a> sin ((theta)/(2)) "" ... (6)` <br/> Since the amplitude of the wave is very small when it is compared with the length of the string , the sine of small angle is approximated as sin `((theta)/(2)) = (theta)/(2)` . Hence equation (6) can be written as `F_(r) = 2 T xx (theta)/(2) = T theta "" .... (7)` <br/> But `theta = (dl)/(R)` ,, therefore substituting in equation (7) , we get <br/> `F_(r) = T (dl)/(R) "" ... (8)` <br/> Applying Newton.s <a href="https://interviewquestions.tuteehub.com/tag/second-1197322" style="font-weight:bold;" target="_blank" title="Click to know more about SECOND">SECOND</a> law to the elemental string in the radial direction, under <a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a> , the radial component of the force is equal to the centripetal force. Hence equating equation (5) and equation (8) , we have `T(dl)/(R) = mu v^(2) (dl)/(R)` <br/> `v = sqrt((T)/(mu))` measured in `ms^(-1) "" ... (9)`</body></html>


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