10 transverse sinusoidal wave is generated at one end of a...

'10) transverse sinusoidal wave is generated at one end of a long, horizontal string by a oscillator moving up and down through a total distance of 1.00 cm: The motion is continuous at a frequency of 120 Hz. The string has linear density of 120 g m and is kept at a tension of F = 90.0 N_ Calculate the maximum value of the transverse speed Calculate the maximum value of the transverse component of tension f Show that the two maximum values found in (a) and (b) occur at the same phase values for the wave, and determine the transverse displacement y at this point:'

Transcript

00:01 In the given problem we have a transverse sinusoidal wave which is generated at one end of a long horizontal string by an oscillator moving up and down through a total distance of one centimeter.

00:11 Now this motion is continuous at a frequency f is equal to 120 hertz and it has a linear density of 120 grams per meter.

00:21 So we have mu is equal to 120 grams per meter.

00:24 So converting this into kilograms we can write 120 times 10 days to minus 3 kilograms per meter.

00:30 Or it can be written as 0 .1 to kg per meter and it is kept at a tension f is equal to 90 newton so we have to calculate the maximum value of the transfer speed and the maximum value of the transverse component of the tension f and we have to show that the two maximum values occur at the same phase value for the wave and we have to determine the transverse displacement y at this point so we know that the wave equation is given by the equation y is equal to a sign kx minus omega t here a is the amplitude of the wave k is a wave number x is a displacement omega is the angular frequency and t is a time taken so the transverse velocity can be found by taking differentiation of this that is d y by d t so this will be equal to d by d y of a sign k x minus omega t so this will give us the transverse velocity u and this will be equal to omega a cos k x minus omega t now the wave number k is given by the expression omega under root mu upon f so we'll substitute it in place of omega so this becomes u equals to omega a cause and this is k is equal to omega under root mu upon f x minus omega t but for the transverse velocity to be maximum we need cost of this whole term to be equal to 0 so our mu max will be equal to omega a because cost of 0 gives us 1 so this whole thing will be equal to 0 1 omega is the angular frequency hence we can write instead of omega 2 pi f so now we will substitute the values we have 2 times 3 .142 times the frequency is 120 hertz and the amplitude here will be the half of the total displacement so this will be d upon 2 which will give us 0 .5 centimeters and if we convert it into meters we will multiply it by 10 raise to minus 2 so this gives us the maximum velocity u max equal to 3 .77 meters per second this is part a of our answer now we have to find the transverse component of the tension so the tension let it be f so this will be trans the transverse component will be sign of the angle phi and we need to first calculate this angle phi so we know that 10 phi is equal to d y by d x which is the slope so using these two equations we will first find the angle so tant phi will be equal to d y by d x but this will be maximum 5 will be maximum when this slope is maximum so this will be equal to d by d x so it will be maximum and instead of y we can write the equation a sign kx minus omega t so this gives us 10 5 max is equal to k a cos k x minus omega t so this will be maximum when our cause of this term will be equal to so this term will be equal to 0...

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