## A transverse sinusoidal wave is moving along a string in the positive direction of an x-axis with a speed of 87 m/s. At t = 0, the string particle at x = 0 has a transverse displacement of 4.3 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = 0 is 16 m/s

### Question:

A transverse sinusoidal wave is moving along a string in the positive direction of an x-axis with a speed of 87 m/s. At t = 0, the string particle at x = 0 has a transverse displacement of 4.3 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = 0 is 16 m/s

(a) What is the frequency of the wave?

(b) What is the wavelength of the wave?

If the wave equation is of the form y(x,t)=ymsin(kx±ωt+ϕ), what are

(c) ym

(d) k

(e) ω

(f) ϕ and

(g) the correct choice

### Answer and Explanation:

Solution

Y=4. cmat x=0, t=0 v=80m/s

Vm=16m/s

**(a)** So the frequency of the wave is f= ohms/2pi=63.66Hz

**(b)** The wave length is ohms = v/f=80/63.66=1.25m/s

**(c) **From the form of the wave equation

Y= ymsin(kr+- wt+radian)

Substituting y=4cm at x=0, t=0

We get ym=0.04m

**(d) **The wave number k is k=2pi/lambda= 2pi/1.25= 5rad/s

**(e)** The angular velocity is given by the relation

ohms=um/ym=16/0.04=400rad/s

**(f)** The phase of the wave function using the condition given before which is

Y=4cmat x=0, t=0 Therefore

0.04= 0.04sin (5(0) +- 400(t) + radian

Solving for sin(radian) =1

Radian=pi/2,

Y=0.04sin(5x + _+400(t) +pi/2)

**(g)** The correct choice of sign in front of ohms is Minus because the wave is shown traveling to positive direction of x-axis

(a) f=63.66Hz

(b) lambda =1.25m/s

(c) ym = 0.04m

(e) ohms = 400rad/s

(f) radian = pi/2, Y = 0.04sin(5x+ -+400(t) + pi/2)

(g) The correct choice of sign in front of ohm is -Ve

## Answer ( 1 )

Answer Above.