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The lowest or fundamental frequency for a stringed instrument is given below:

f 0 = 1/2l √T/m

f 0 = fundamental or lowest frequency (Hz)
l = length of string (m)
T = tension force in string (N)
m= mass of 1m of the string (kg)

On a particular piano two adjacent low notes use the same tension in the wire of mass per unit length of 0.030 kg m-1. One of the wires is 0.78m long and it vibrates at a frequency of 55Hz. What is the tension in the string? The other string is slightly longer at 0.91m. What is the frequency of its vibration?
2. (Original post by a5h)

The lowest or fundamental frequency for a stringed instrument is given below:

f 0 = 1/2l √T/m

f 0 = fundamental or lowest frequency (Hz)
l = length of string (m)
T = tension force in string (N)
m= mass of 1m of the string (kg)

On a particular piano two adjacent low notes use the same tension in the wire of mass per unit length of 0.030 kg m-1. One of the wires is 0.78m long and it vibrates at a frequency of 55Hz. What is the tension in the string? The other string is slightly longer at 0.91m. What is the frequency of its vibration?
Its a simple question
f0 = ½l √T/ρ (I prefer using ρ to m since its more of a density than mass)
f0 = 55Hz = ½ 0.78 √T/0.030 kg m-1
110 / 0.78 = √T/0.030 kg m-1
141.026 = √T/0.030 kg m-1
19888 = T/0.030 kg m-1
19888 x 0.030 kg m-1 = T
T = 596.647kg (technically the wrong unit for tension but who gives)

Then we need to compare f' with f.
Since f = ½l √T/ρ
and the only change in the string is l.
therefore we can reduce the formula to f = l k (where k = ½√T/ρ although this is unimportant)
Hence we work out k for the first string. 55Hz = 0.78m k
k = 70.512Hz m-1.
Then for
f' = 0.91m k
f' = 0.91m 70.512Hz m-1
f' = 64.167Hz

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Updated: July 6, 2005
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