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 one 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 with 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, μ is
The elemental string AB has a curvature which looks like an arc of a circle with centre at O, radius R and the arc subtending an angle θ at the origin O. The angle θ can be written in terms of arc length and radius as θ = \(\frac{dl}{R}.\)
The centripetal acceleration supplied by the tension in the string is
acp = \(\frac{v^2}{R}\) ....(3)
Then, centripetal force can be obtained when mass of the string (dm) is included in equation (3)
Fcp = \(\frac{(dm)v^2}{R}\) ....(4)
The centripetal force experienced by elemental string can be calculated by substituting equation (2) in equation (4) we get
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 (\(\frac{θ}{2}\)) and vertical component T sin (\(\frac{θ}{2}\)) The horizontal component at A and B are equal in magnitude but opposite 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 Fr is
Since the amplitude of the wave is very small when it is compared with the length of the spring, the sine of small angle is approximated as sin (\(\frac{θ}{2}\)) ≈ \(\frac{θ}{2}\). Hence equation (6) can be written as
Applying Newton’s second law to the elemental string in the radial direction, under equilibrium, the radial component of the force is equal to the centripetal force. Hence equating equation (5) and equation (8), we have
