Show which formulas are used when solving. 1. A string vibrates according to the

Question

Show which formulas are used when solving.

1. A string vibrates according to the equation Y(x,t) = 700 cm * sin((5.00cm-1)x)*cos(8.00 rad/s)t (THIS IS A STANDING WAVE.) What is the angular wave number? What is the wavelength? What is the speed of the 2 waves making the standing wave? What is the amplitude of the 2 waves making the standing wave? What is the distance between the nodes? What is the angular speed of the 2 waves making the standing wave? What is the period of the 2 waves making the standing wave? What is the maximum transverse speed of this standing wave? Write the equations for the two waves making up the standing wave Y 1 (x,t) = Y2(x,t) =

Explanation / Answer

Given

standing wave equation Y(x,t) = 700 cm sin(5.00cm^-1)x *cos(8.00 rad/s)t

we know that a standing wave will form when two waves are of same amplitude frequency and are propagating in opposite directions then superimpose forms a standing wave

whose amplitude will be double the amplitude of individual wave

that is  

y1(x,t) = A sin (kx-wt) moving in the +ve x direction

y2(x,t) = A sin (kx+wt) moving in the -ve x direction

superposition of the waves is

Y(x,t) = 2A sin (kx)*cos (wt)

that is sin (A-B) = sin A cos B - cos A sin B

sin (A+B) =sin A cos B + cos A sin B

adding these sin (A+B)+sin(A-B) = 2 sin A cos B

y(x,t) = y1(x,t)+y2(x,t) = 2A sin (kx) cos(wt)

so k is wave number , w is angular velocity  

k = 2pi/lambda = 5 cm ==> lambda = 2pi/5 cm = 1.2566 cm

angular number k = 5 cm^-1

wavelength lambda = 1.2566 cm

wave speed is v = lambda*f = lambda*2pi/w = 1.2566*2pi/8 cm/s = 0.9869 cm/s

amplitude of the standing wave is 2A = 700 cm = 7 m

Distance between the nodes is = lambda/2 = 1.2566 /2 = 0.6283 cm

Angular speed is W = 8 rad/s

period of the wave is W = 2pi/T ==> T = 2pi/W = 2pi/8 a = 0.7853 s

y1(x,t) = 350cm sin (5cm)x -8t)

y2(x,t) = 350 cm sin (5 cm + 8t)

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