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

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

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Sam Professor 2 years 1 Answer 1062 views 0

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