Solving question of progressive wave Equations

welcome to today's class today i'm going to look at another concept in physics under with now what to this concept you're going to look at the progressive with equations and then the concept you to prove these equations for instance you have a progress with with displacements y i'm moving three distance x and the equation can be given as y equal to a sine open bracket wt minus 5 call this equation 1 y represents vertical displacements a represents amplitude of the wave w represents angular frequency y t represents time taken and this is face angle that is where y represents displacements a represents what amplitude this sign represents angular frequency t represent time taking these five represents phase angle but record that in simple harmonic motion angular frequency is given as 2 pi f where f represents frequency of the wave y first angle is given as 2 pi x over lambda now let's substitute the value that is in equation now in equation one we are going to substitute the value of omega angular frequency and phi therefore the equation will turn to y equal to a sine open brackets 2 pi f t minus 2 pi x over lambda now look at this equation there is common factors the common factor here is 2 pi now factorize the equations and then we have y not equal to a sine two pi open bracket f t minus x over lambda now record that under the wave equations recall that v is equal to lambda f that is wavelength and frequency now make f the subject of the equations therefore we have f equal to v over lambda now go back to this equation remove this f and substitute this value there therefore we have y equal to a sine 2 pi v t over lambda minus x over lambda also 1 over lambda is a common factor of the equation therefore y equal to a sine 2 pi over lambda into vt minus x as the equation now from this angle now where v is the wave speed x is the horizontal displacement of the wave particle okay now you compare and contrast this equation of progressing with y is equal to a sine open bracket with t minus five now we have y go twice therefore the amplitude amplitude equal to 0.4 meter and then angular frequency is 115. take this value equality to this the graph wt that was angular frequency equal to 150 t t go with t the angular frequency equal to 150 we remember that angular frequency is given as 2 pi f that is 2 pi f equal to 150 now make f the subject of the equation this 2 pi 2 pi f not equal to 75 over pi and then 75 over pi equal to 20 0.78 x and then the next value obtain our frequency and then remember that the frequency and period the inverse of each order now is the inverse of each other now that means that period equal to one over f which is equal to one over two series seven eight the equal to zero points equal to zero point zero four two there's zero point zero four two second now that's value of of period and then the next value is wavelength how are you going to obtain the wavelength and you can obtain the wavelength from this side that is you take this value now 5 that is phi equal to 200 x and that was 5 5 what was 2 pi x over lambda equal to 200 x x go with x that's two pi over lambda is equal to two hundred to go here one that's one hundred now pi over five sorry pi over lambda equal to 100 therefore lambda is equal to pi over 100 what is pi pi is 3.142 over 100 is equal to what 0.03142 meter does pi and angular sorry wavelength now to get the wave speed and then record the equation that's the third one d v equal to lambda f we are lambda is wavelength f is the frequency therefore we have v not equal to 0.03 one four two multiplied by the frequency that's 23.78 now when you multiply this value together it will give you 0.75 v 9 is equal to 0.75 meter per second

Channel: Brainstorm Learning Center

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Id: TST-2ffSAuY

Length: 8min 52sec (532 seconds)

Published: Mon Jun 29 2020