The transverse displacement of a wire (clamped at both its ends) is described as : y(x, t)=0.06 \sin \left(\frac{2 \pi}{3} x\right) \cos (120 \pi t) The mass of the wire is 6 \times 10^{-2} \mathrm{~kg} and its length is 3 \mathrm{~m}.
Provide answers to the following question:
Calculate the wire’s tension.
[X and y are in meters and t in secs]
The transverse displacement of a wire (clamped at both its ends) is described as : y(x, t)=0.06 \sin \left(\frac{2 \pi}{3} x\right) \cos (120 \pi t) The mass of the wire is 6 \times 10^{-2} \mathrm{~kg} and its length is 3 \mathrm{~m}.
Provide answers to the following question:
Calculate the wire’s tension.
[X and y are in meters and t in secs]

As we know,

The standard equation of a stationary wave is known as,

y(x, t)=2 a \sin k x \cos w t

Given equation is,

y(x, t)=0.06 \sin \left(\frac{2 \pi}{3} x\right) \cos (120 \pi t)

It is similar to the general equation .

Velovity of transverse wave is given as v=180m/s

Mass of the string is given as m=6\times10^{-2}kg

String length is given as I=3m

Mass per unit length can be calculated as,

\mu=m/l=\frac{6\times10^{-2}}{3}
=2\times10^{-2}kg/m

Let T be the tension in the wire.
So, T=v^{2}\mu
=180\times180\times2\times10^{-2}
=648N