Question

4) A piano tuner hits a key on a piano and strikes a 320 Hz tuning fork. a) The tuner hears a beat frequency of 5 Hz. What could the piano frequency be? Explain. b) The tuner then tightens the piano string which increases its frequency. When he does, the beat frequency increases. What is the original piano frequency? Explain. 5) In a 0.5 m tube, a standing sound wave is created at 857.5 Hz. The speed of sound is 343 m/s. a) Draw a picture of the wave in the tube. b) Which harmonic is the sound? c) Are the ends of the tube open or closed? A) Both ends are open B) Both ends are closed C) 1 end is open, the other is closed d) What is the fundamental frequency of the tube?

Answer 1

4)

a) f_beat = | f1 – f2|

5 = |320 – f2|

Hence f2= 315 or 325

b) Original frequency of the piano must be f1= 325Hz , because increasing f2=325Hz increases beat frequency.

5)L= 0.5m

v =343m/s

f= 857.5 Hz

a)

b)f= (5*343)/(4*0.5) = 857.5 Hz

Comparing this equation with standard equation,

f=(nv)/(4L)…………………. for closed tube at one end,

we get,

n=5

Thus the harmonic is 5^th

c) Tube is closed at one end

d) For fundamental frequency n= 1

f=(nv)/(4L)= (1*343)/(4*0.5) = 171.5 Hz