Question

Vibration from a 620 Hz tuning fork sets up standing waves on a string clamped at both ends (one of which you may take as x=0). The travelling wave speed on this string is 440 m/s. The standing wave has four anti-nodes and an amplitude of 2.0 mm.

(a) What is the length of the string?
___ m

(b) Write an equation for the transverse displacement of the string as a function of position and time:

y(x,t) = ( ___ mm) sin ( ___ m^-1 x ) cos ( ___ s^-1 t)

(c) What is the wavelength of the travelling waves that form this standing wave?
___ m

Answer 1

Given that :

frequency, f = 620 Hz

speed of travelling wave, v = 440 m/s

maximum amplitude, A_max = 2 mm

The standing wave has four anti-nodes.

(a) the length of the string will be given as ::

using an equation, f = n v / 2 L

Or   L = n v / 2 f                                                                           { eq.1 }

where, n = 4

inserting the values in eq.1,

L = (4) (440 m/s) / 2 (620 Hz)

L = (1760 m/s) / (1240 Hz)

L = 1.41 m

(b) An equation for the transverse displacement of the string as a function of position and time which is given as :

y (x,t) = A_max Sin (kx) . Cos t                                                                  { eq.2 }

where, = 2f = 2 (3.14) (620 Hz)

= 3893.6 rad/s

and k = / v = (3893.6 rad/s) / (440 m/s)

k = 8.85 m^-1

Now, inserting these values in eq.2 -

y (x,t) = (2 mm) Sin [(8.85 m^-1) x] . Cos [(3893.6 s^-1) t]                 

(c) The wavelength of the travelling waves that form this standing wave which will be given as ::

= 2 / k                                                                                      { eq.3 }

inserting the values in eq.3,

= 2 (3.14) / (8.85 m^-1)

= (6.28) / (8.85 m^-1)

= 0.709 m