Consider a loop in the standing wave created by two waves...

Consider a loop in the standing wave created by two waves (amplitude 5.00 mm and frequency 120 Hz) traveling in opposite directions along a string with length 2.25 m and mass 125 g and under tension 40 N. At what rate does energy enter the loop from (a) each side and (b) both sides? (c) What is the maximum kinetic energy of the string in the loop during its oscillation?

Transcript

00:01 Hello everyone so in this motion we can divide the answer in the steps like moving to the first step the mass per unit length of the string is mu that is m by m divided by l now putting the value 0 .125 divided by 2 .25 divided by 2 .25 and if i see this i get 0 .0555 5 kg per meter and the speed to on the string is b that is root of tension divided by mu and that mu i have calculated before and the tension i know from the question that is 40 that is 0 .05 if i see and calculated this answer to be 26 .83 meter per second so this is the velocity okay now going to the step two and seeing the angular speed on the string omega this is angular speed and this would be equal to 2 pi f where i feel the frequency and 2 multiply by 5 3 .14 or 27 by 8 1 2 double 0 and if i see like 7 this would be equals to 7 and that 5 54 080 radium per second okay now this i think point something is different so this is having 7540 .8 okay so this is our angular speed now mind height and seeing the rate at which the energy is transmitted by a sinusoidal wave on a string is given as power equals to average power that is given divided by 2 mu a frequency omega square m okay now putting all these value 1 by 2 mu 0 .055 and new 26 .83 multiply by 75408 whole square and 0 .05 and 0 .055 and 0 .055 whole square and 0 .055 okay, now the cycle greater then it will be equal to now move to the step three sorry that the rate at which the energy enters the loop from both sides is 2 multiply by 3 of average that is equal to 2 multiplied by 10 .5 9 and this will be equals to 21 .2 watt coming at and seeing the differentiation it is dk by d t and this is equal to half of mu mu omega square y square coarse square kx minus omega p your sinusoidal for both side k is equal to muon course kx minus omega t d t okay now if i integrate this then this will be out of course k x this is sparsely k x minus omega t by 2 this is my k now this k is also can be written x t average multiplied by t by 2 and k is equal to minus t average divided by 2 f so t can be going to this tab for substituting the given value k in regards to the average that is 21 .19 and 2 times a 120 and this is equal to 0 .083 and k final equals to 8 .8 3 multiply by 10 raised to 4 minus 2 2 so this is for k maximum kinetic energy for the this is our answer...

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