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For the wave discribed in Exercise 15.8, plot the displacement (y) versus (t) graphs for x=0, 2 and 4 cm. What are the shapes of these graphs ? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase ? |
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Answer» Solution :Given wave EQUATION is `y = 3 sin (36 t +0.018 x + (pi)/(4)) cm ""…(1)` `implies` Phase of a particle at distance x from origin of wave at time t is `theta = 36 t + 0.018x + (pi)/(4) rad ""…(2)` Now we are given three particles whose displacement are to be found out at different instnats of time. They are LOCATED at `x=0,` x=2 cm and x=4 cm respectively. Let us consider their simple harmonic oscillations one by one. For particle at `x=0,` from equation (1), `y =3 sin (36 t + (pi)/(4)) ""...(3)` `therefore y= a sin ((2pi )/(T) t + (pi)/(4))` (Where `(2pi)/(T) = omega = 36 (rad)/(s)) ...(4)` From above equation we can find out vaues of DISPLACEMENTS y at `t =0, (T)/(8), (2T)/(8), (3T)/(8), (4T)/(8),...,` `(8T)/(8) = T` as flollows. `(1) At t =0, y = a sin "" (pi)/(4) = (a)/(sqrt2)` (2)` At t = (T)/(8), y = a sin ((2pi)/(T) xx (T)/(8) + (pi )/(4))` `=a sin ((pi)/(2)) =a` `(3) At t = (2T)/(8), y =a sin ((2pi)/(T) xx (2T)/(8) + (pi)/(4))` `=a sin ((pi)/(2) +(pi)/(4)) =a cos ""(pi)/(4) = (a)/(sqrt2)` `(4) At t= (3T)/(8), y= a sin ((2pi)/(T) xx (3T)/(8) + (pi)/(4))` `=a sin (pi) =0` `(5) At t = (4T)/(8), y = a sin ((2pi)/(T) xx (4T)/(8) + (pi)/(4))` `= a sin (pi + (pi)/(4)) =a (-sin "" (pi)/(4)) =- (a)/(sqrt2)` `(6)At t = (5t)/(8), y = a sin ((2pi)/(T) xx (5T)/(8) + (pi)/(4))` `= a sin ((3PI)/(2)) =-a ` `(7) At t = (6T)/(t), y = a sin ((2pi )/(T) xx (6T)/(8) + (pi)/(4))` `= a sin ((7pi)/(4)) =a sin (2pi - (pi)/(4)) =- (a)/(sqrt2)` `(8) At t = (7T)/(8), y = a sin ((2pi)/(T) xx (7T)/(8) + (pi)/(4))` `= a sin (2pi) =0` `(9) A t t = (8T)/(8) =T,y = a sin ((2pi )/(T) xx T + (pi)/(4))` `=a sin ((pi)/(4)) = (a)/(sqrt2) ` From above calculations, graph of `y to t` can be plotted as follows :  At `t =0,` initial phase of a particle at distance x is [from equation (2)], `theta = 0.018 x + (pi)/(4)` `implies ` For first particle, `x = 0 implies theta _(1) = (pi)/(4) rad` For SECOND particle, `x =4 cm implies theta_(3) =0.072 + (pi)/(4) rad` For all of above particles wa can draw graph of `y to t ` as shown earlier. For all of them, amplitude, frequency and wavelength are same, but their initial phase are different. |
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