Question

The transverse displacement of a string (clamped at its both ends) is given by y(x,t)=0.06 sin (2π/3x) cos (120πt) where x and y are in m and tin s. the length of the string is 1.5 m and its mass is 3.0×10² kg. Determine the tension in the string.

Answer 1

The transverse displacement of a string is given by y(x,t) = 0.06sin(2π/3x)cos(120πt).

we know, 2sinA.cosB = sin(A + B) + sin(A - B)
use this formula here to separate wave equation into two waves .

e.g., y(x,t) = 0.03[2sin(2π/3x)cos(120πt)]
= 0.03[sin(2π/3x + 120πt) + sin(2π/3x - 120πt)]
= 0.03[sin(120πt + 2πx/3) - sin(120πt - 2πx/3)]
we two progressive waves are
y1 = 0.03sin(120πt + 2πx/3)
y2 = 0.03sin(120π - 2πx/3)

now compare it with standard equation of progressive wave e.g., $y=asin(\omega t\pm kx)$

we get, a = 0.03m or 3cm
e.g., amplitude = 3cm

$\omega$ = 120π rad/s
2πf = 120π rad/s
f = 120π/2π Hz = 60Hz
hence, frequency = 60Hz

k = 2π/3
2π/wavelength = 2π/3
wavelength = 3m

so, speed of wave = wavelength × frequency
= 3 × 60 = 180m/s

now, tension in string is given by , T = v² m
where m is mass per unit length and v is the speed of wave.

Given, mass of string is 3 × 10^-2 kg
length of string is 1.5m
so, mass per unit length , m = 3 × 10^-2/1.5
= 2 × 10^-2 kg/m
speed is sound , v = 180

now, T = (180)² × 2 × 10^-2
= 32400 × 2 × 10^-2
= 648N