Question

The function y(z, t)-(0.150 m) cos (та-15nt) where æ is in meters, and t is in seconds, describes a transverse wave on a taught string. a) What is the transverse velocity of a point on the string at an instant in time when y- +0.120 m? b) What is the maximum transverse velocity a point on the string can have? c) What is the average transverse velocity of a point on the string averaged over an integer number of wave periods?

Answer 1

y = 0.15 cos(pi x - 15 pi t)

at y = 0.12 ;

0.12 = 0.15 cos@

@ = 36.87 deg

Vy = 0.15 x 15 pi sin (pi x - 15 pi t)

= 0.15 x 15 pi x sin@

= 4.24 m/s

Vmax = 0.15 x 15 pi = 7.065 m/s

Vavg = 0 ... because wave

V = wavelength / T

but wavelength = 2 and T = 2/15 ... comparing with standard eqn

V = 2/(2/15) = 15 m/s

No comparison. wave speed is constant. Transverse speed changes with time and position