(30) A stretched string of length L is observed to vibrate in six equal segments
ID: 2255554 • Letter: #
Question
(30) A stretched string of length L is observed to vibrate in six equal segments when driven by a 725-Hz oscillator. What oscillator frequency will set up a standing wave so that the string vibrates in two segments? answer in Hz
(29)How far, and in what direction, should a cellist move her finger to adjust a string's tone from an out-of-tune 447 Hz to an in-tune 440 Hz? The string is 68.0 cm long, and the finger is 17.9 cm from the nut for the 447-Hz tone. (Indicate the direction with the sign of your answer. Take the direction towards the nut to be positive.)
cm
N
Explanation / Answer
30)
since velocity in string remains same coz tension remains same
?1xf1=f2x?2
?1=2L/5
?2=L
f1=725
we get f2=290 Hz
29)
Use the formula v = f?
The velocity for bot notes will be the same so we can set up a proportionality
f1?1 = f2?2
The wavelength will be twice the string vibrating length
The string vibrating length is .68 - .179 = .501, so the wavelength is 1.002 m
(447)(1.002) = (440)(?2)
?2 = 1.01794 m
String length of vibration must be .50897 cm
Since the total string lenght is .68, then .68 - .50897 = .17103 m or 17.103 cm
Since her finger is 18.6 cm from the nut, she needs to move her finger toward the nut. The distance is
18.6 - 17.103 = 1.497 cm
28)
f = n v / 2 L
v = ?T/?
? = 3.7 x 10-3 / 0.5
= 7.4 x10-3
261.6 = 1 x ?(T/ 7.4 x10-3) / 2 x 0.5
T = 506.415 N
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