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Two sinusoidal waves combining in a medium are described by the wave functions y_(1) = ( 3.0 cm) sin pi ( x + 0.60 t) y_(2) = ( 3.0 cm) sin pi ( x - 0.60 t) where x is in centimeters and t is in seconds . Determine the maximum transverse position of an element of the medium at (a)x = 0.250 cm , (b) x = 0.500 cm and (c ) x = 1.50 cm. (d) Find the three smallest values of x corresponding to antinodes. |
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Answer» Solution :The answers for maximum transverse position must be between ` 6 cm and 0`. The anitodes are seperated by half a wavelength , which we except to be a COUPLE of centimetres . According to the WAVES in interference model , we write the function ` y_(1) + y_(2)` and start evaluating things. We get ` y = y_(1) + y_(2) = ( 6.0 cm) sin ( pi x) cos ( 0.60 pi t)` SINCE ` cos (0) = 1` , we can find the minimum value of `y` by SETTING ` t = 0` . `y_(MAX) (x) = y_(1) + y_(2) = ( 6.0 cm) sin (pi x)` (a) At `x = 0.250 cm , y_(max) = ( 6.0 cm) sin ( 0.250 pi) = 4.24 cm` (b) At ` x= 0.50 cm , y_(max) = ( 6.0 cm) sin ( 0.500 pi ) = 6.00 cm` ( c) At ` x= 1.50 cm , y _(max) = | ( 6.0 cm) sin (1.50 pi)| = + 6.00 cm` (d). The antinodes occur when ` x = n lambda //4 for n = 1 , 3 , 5 , .....` But ` k = 2 pi //lambda = pi , so lambda = 2.00 cm ` and ` x_(1)= lambda//4 = ( 2.00 cm)//4 = 0.500 cm` `x_(2) = 3 lambda //4 = 3 ( 2.00 cm )//4 = 1.50 cm` ` x_(3) = 5 lambda //4 = 5 ( 2.00 cm)//4 = 2.50 cm` Two of our answers in (d) can be read from the way the amplitude had its largest possiblr value in parts (b) and ( c ). Note again that an amplitude is defined to be always positive , as the maximum absolute value of wave function . |
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