A very long string with 20 gm is stretched along the x axis...

A very long string with Î¼ = 2.0 g/m is stretched along the x-axis with a tension of 5.0 N. At x = 0 m, it is tied to a 100 Hz simple harmonic oscillator that vibrates perpendicular to the string with an amplitude of 2.0 mm. The oscillator is at its maximum positive displacement at t = 0 s. a. Write the displacement equation for the traveling wave on the string. b. At t = 5.0 ms, what is the string's displacement at a point 2.7 m from the oscillator?

Transcript

00:01 Hello students in this question we are given that a linear density of the rope the linear density which is represented by mu is equal to 2 multiplied with 10 to the power minus 3 kg per meter and the tension in the rope is equals to 5 newtons and the frequency to which this rope is connected is hundred hertz the amplitude is given to us as 2 multiplied with 10 to the power minus 3 meters now here first of all we will calculate the velocity of the wave on the string so it will be equals to so firstly we will write our formula for the velocity so velocity is equal to square root of t divided by mu in a stretched string so it will be equal to square root of 5 over 2 multiplied with 10 to 1 minus 3.

01:13 So it will be 50 meters per second.

01:17 Now the wave number which is k, wave number which is k is equal to w divided by v.

01:35 So w is nothing but 2 pi f divided by v.

01:42 So it will be 2 multiplied by pi multiplied by 100 divided by v which is 50 so it will be 12 .56 newton per meter now we have calculated the wave number now the angular frequency is omega so it will be equals to 2 pi f so it will be 2 pi f so it will be 2 pi f so it will be equals to 2 pi multiplied by f which is 100 so it will be 628 radian per second now in the first part we have to write the displacement equation so our displacement equation y is equal to a sine k x minus omega t so this is our function that is why we have found out these values so y will be equals to a which is 2 multiplied with 10 to the power minus 3 multiplied with sign of k k is 12 .56 x minus value of omega is 628 t so this is the displacement equation now in the b part we have to find that time equals to 5 milliseconds time is given to us as 5 milliseconds we have to find a displacement at a point which is x equals to 2 .7 meters.

03:26 So now y will be equals to 2 multiplied with 10 to the power minus 3.

03:32 Sign of 12 .56 multiplied by x which is 2 .7 minus 628 t which is 5 multiplied with 10 to the power minus 3...

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