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!
frequency new=frequency old *sqrt(169/289)
http://hyperphysics.phy-astr.gsu.edu/Hbase/waves/string.html
thanks Bob. Can you please help me with question A as well?
To find the frequency in each situation, you can use the formula for the frequency of a vibrating string:
f = (1/2L) * sqrt(T/μ)
Where:
- f is the frequency in Hz
- L is the length of the string in meters
- T is the tension in Newtons
- μ is the linear density of the string in kg/m (μ = m/L, where m is the mass of the string)
Let's calculate the frequency in each situation:
a) When the tension is reduced to 196 N:
- The length of the string (L) remains the same at 60 cm = 0.6 m
- The diameter of the string is also unchanged at 1.4 mm = 0.0014 m
- The linear density (μ) of the string can be calculated using the mass (m) and length (L):
- To find the mass of the string, we need the density (ρ) and volume (V)
- The density of steel, for example, is around 7850 kg/m^3
- The volume of a cylinder is given by V = πr^2L, where r is the radius (diameter/2)
- Plugging in the values: V = π(0.0007)^2 * 0.6.
- Now, the mass of the string can be found by multiplying the volume by the density: m = ρV = 7850 * V.
- Finally, the linear density is μ = m/L.
Once you have μ and T, you can calculate the new frequency using the formula mentioned earlier.
b) When a string of the same material, 45 cm long and 1 mm in diameter, is plucked under 168 N of tension:
- The length of the string (L) is now 45 cm = 0.45 m.
- The diameter of the string is 1 mm = 0.001 m.
- Similar to the previous case, you need to calculate the linear density (μ) for this string, as described before.
Once you have the μ value for the second string, you can use the formula to calculate the new frequency.
I hope this helps! Let me know if you have any further questions.