### Q.11:- The transverse displacement of a string (clamped at its both ends) is given by

### Where *x *and *y *are in m and *t *in s. The length of the string is 1.5 m and its mass is 3.0 ×10^{–2} kg.

Answer the following:

(a) Does the function represent a travelling wave or a stationary wave?

(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

(c) Determine the tension in the string.

**Answer:-**(a)** **The general equation representing a stationary wave is given by the displacement function:

*y* (*x*, *t*) = 2*a* sin *kx *cos ω*t*

This equation is similar to the given equation:

Hence, the given equation represents a stationary wave.

(b) A wave travelling along the positive

*x*-direction is given as:*y*

_{1}=

*a*sin (

*ωt*–

*kx*)

The wave travelling along the positive

*x*-direction is given as:*y*

_{2}=

*a*sin (

*ωt +*

*kx*)

The supposition of these two waves yields:

*y*=

*y*

_{1 + }

*y*

_{2 = }

*a*sin (

*ωt*–

*kx*) –

*a*sin (

*ωt +*

*kx*)

=

*a*sin (*ωt*) cos (*kx*) –*a*sin (*kx*) cos (*ωt*) –*a*sin (*ωt*) cos (*kx*) –*a*sin (*kx*) cos (*ωt*)= 2

*a*sin (*kx*) cos (*ωt*)∴Wavelength, *λ* = 3 m

It is given that:

120π = 2πν

Frequency, * ν* = 60 Hz

Wave speed,

*v*=

**ν**λ= 60 × 3 = 180 m/s

(c)** **The velocity of a transverse wave travelling in a string is given by the relation:

*v* = √*T*/*µ* ….**(i)**

where,

Velocity of the transverse wave, *v* = 180 m/s

Mass of the string, *m* = 3.0 × 10^{–2} kg

Length of the string, *l *= 1.5 m

Mass per unit length of the string, *µ = m/l*

= 3.0 × 1.5 = 10^{-2}

= 2 × 10^{-2} kg m^{-1}

Tension in the string =

From equation (

= (180)

= 648 N

*T*From equation (

*i*), tension can be obtained as:*T*=*v*^{2}*μ*= (180)

^{2}× 2 × 10^{–2}= 648 N