(i) For the wave on a string described in Exercise 15.11. do all the points on the string oscillate with the same (a) frequency , (b) phase , (c ) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?

(i) For the wave on a string described in Exercise 15.11. do all the p

1.	(i) For the wave on a string described in Exercise 15.11. do all the points on the string oscillate with the same (a) frequency , (b) phase , (c ) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?
Answer» Solution :Here given stationary wave is `y = 0.06 SIN ((2pi)/(3)x) cos (120 pi t ) ""…(1)` Comparing above equation with, `y = 2a sin (kx) cos (omegat)""…(2)` we get ,` omega 2pi f = 120 pi impliesf = 60 Hz` which is same for all the particles of a string. In case of stationary wave, PHASES of oscillation for particles in any given loop are same because they oscillate simulataneously PERPENDICULAR to string. (In other words, phase differentce between any two particles in any oop is ZERO). In case of stationary wave AMPLITUDES of all the particles in a given string are not same because amplitude of oscillation of a given particle at distnace x from one end of string is, `A =2a sin (kx)` `implies` Amplitude of oscillation of a particle depends on its position. Here,` A = 3 a sin (kx)` `= 0.06 sin ((2pi)/(3) xx 0.375) = 0.067 sin (0.25pi)` `=0.06 sin ((pi)/(4))` `=0.06 xx 0.7071` `therefore A = 0.04343 m`