Asked by Sarah N H
A guitar string 60 cm in length, with a diameter of 1.4 mm and a tension of 289 N, emits a note with a frequence of 147 Hz. Find the frequency in each of the followig situations:
a)the tension isreduced to 196N
b)a string of the same material, 45 cm long and 1mm in diameter under 168 N of tension, is plucked.
a)I have no idea for a..
b)I was hoping someone could check my work on b?
=Fi(square root of tension fraction)(length fraction)(diameter fraction)
=147Hz(square root of 169/289)(60/45)(1.4/1)
=209.8Hz
I am just getting confused as to if the 289 goes on top of the fraction, or the 169.
thanks for your help in advance!
a)the tension isreduced to 196N
b)a string of the same material, 45 cm long and 1mm in diameter under 168 N of tension, is plucked.
a)I have no idea for a..
b)I was hoping someone could check my work on b?
=Fi(square root of tension fraction)(length fraction)(diameter fraction)
=147Hz(square root of 169/289)(60/45)(1.4/1)
=209.8Hz
I am just getting confused as to if the 289 goes on top of the fraction, or the 169.
thanks for your help in advance!
Answers
Answered by
bobpursley
frequency new=frequency old *sqrt(169/289)
http://hyperphysics.phy-astr.gsu.edu/Hbase/waves/string.html
http://hyperphysics.phy-astr.gsu.edu/Hbase/waves/string.html
Answered by
Sarah N H
thanks Bob. Can you please help me with question A as well?
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