points separated by a distance of ( 10 mathrm{~m} ) along ( mathrm{x} ) - axis is ( X pi ) then ( mathrm{x} ) is 0.5 2. Transverse wave is propagating in a string. Tension in the string becomes twice to the initial tension. Simultaneously, area of cross-action of the string is increased so that there is no change in speed of the wave. Initial cross-section area is ( A_{0} ) Final cross-section area is ( K A_{n} ). Find the value ( mathrm{K} quad 0.5 )

The speed of a transverse wave on a string is 110 m s−1. The tension in the string is340 N. Calculate the mass per unit length of the string

Transverse waves with a speed of 46.5 m/s are to be produced on a stretched string. A 4.30-m length of string with a total mass of 0.0600 kg is used.(a) What is the required tension in the string? N(b) Calculate the wave speed in the string if the tension is 8.00 N. m/s

Calculate the speed of a wave travelling along a stretched string that is under a tension of50 N and has a mass per unit length of 3.110–4 kg m–1

The linear density of a vibrating string is 10−4kg/m. A transverse wave is propagating on the string, which is described by the equation y = 0.02 sin (x + 30 t), where x and y are in meters and time t in seconds. Then tension in the string is

A string of mass 2000g and length 100metres is under tension of 150N. (i)  Determine the linear density of the string (iii) Velocity of wave in the stringa.0.02N/M (ii) 0.866m/sb.0.03N/M (ii) 8.86m/sc.0.2N/M (ii) 0.866m/sd.0.02N/M (ii) 8.00m/se.0.4N/M (ii) 8.00m/sf.0.4N/M (ii) 0.0866m/s