A thin taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x,t) = (5.60 cm) sin [(0.0340 "rad"/cm)x] sin [(50.0 "rad"/s)t], where the origin is at the left end of the string, the x-axis is along the string and the y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing wave pattern. (b)Find the amplitude of the two travelling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period and speed of the travelling wave. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation y(x,t) for this string if it were vibrating in its eighth harmonic?

A thin taut string tied at both ends and oscillating in its third harm

1.	A thin taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x,t) = (5.60 cm) sin [(0.0340 "rad"/cm)x] sin [(50.0 "rad"/s)t], where the origin is at the left end of the string, the x-axis is along the string and the y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing wave pattern. (b)Find the amplitude of the two travelling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period and speed of the travelling wave. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation y(x,t) for this string if it were vibrating in its eighth harmonic?
Answer» Solution :(a) Third harmonic (b) `A= 5.6/2 = 2.8cm ` (c) `k= 0.034 cm^-1` `lambda = (2pi)/k = 184.7 cm ` `L = (3lambda)/2 = 277 cm ` (d) ` lambda = 184.7 cm ` `f = omega/(2pi) = 50/ (2pi) = 7.96 Hz.` `T = 1/f = 0.126 s` ` v= flambda = 1470 cm//s ` (e) `v_(MAX) = omega A_(max)` = `(50)(5.60)` `= 280 cm//s` (f) Frequency and HENCE `omega` will BECOME 8/3 times. `:. omega' = 50 xx 8/3 = 133 rad//s` `k = omega/v` or `k prop omega` Hence, k will become`8/3` times. `k' = (8/3) (0.034) = 0.0907 "rad"// cm .`