Question

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 193 m/s and a frequency of 235 Hz . The amplitude of the standing wave at an antinode is 0.380 cm.

A) Calculate the amplitude at a point on the string a distance of 18.0 cm from the left-hand end of the string.
B) How much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point?
C) Calculate the maximum transverse velocity of the string at this point.
D) Calculate the maximum transverse acceleration of the string at this point.

Answer 1

A. We know that amplitude at x is

Asin (kx)

But k= 2πf/v

k= 2*3.132*235/193= 7.65

So A = 0.35*sin( 7.65x 0.18)= 0.00841m

C

Vmax = Amplitude x angular velocity

= 0.0084 x 2πf

= 0.0084* 2*3.142* 235= 12.4m/s

D. Maximum acceleration = omega² x Amplitude

= (2πf)²* 0.00841= 183.40m/s²