A wire gives a fundamental note of frequency 288 Hz when it has a tension of 36 newtons.
(i) At what tension will the wire emit a note of frequency 432 Hz?
(ii) How would you make the wire emit a note of frequency 720 Hz, keeping the tension at 36 newtons?
Calculate the fundamental frequency of a string 0.5 metre long and weighing 0.004 kg, when stretched by a load of 225 kg.
The Correct Answer and Explanation is :
Solutions
(i) Tension for a frequency of 432 Hz
The fundamental frequency ( f ) of a vibrating string is related to the tension ( T ) by:
[
f \propto \sqrt{T}
]
This implies:
[
\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}}
]
Given:
- ( f_1 = 288 \, \text{Hz} ), ( T_1 = 36 \, \text{N} )
- ( f_2 = 432 \, \text{Hz} ), ( T_2 = ? )
[
\frac{432}{288} = \sqrt{\frac{T_2}{36}}
]
[
\frac{3}{2} = \sqrt{\frac{T_2}{36}}
]
Square both sides:
[
\left(\frac{3}{2}\right)^2 = \frac{T_2}{36}
]
[
\frac{9}{4} = \frac{T_2}{36}
]
[
T_2 = 36 \times \frac{9}{4} = 81 \, \text{N}
]
Answer: ( T_2 = 81 \, \text{N} )
(ii) Frequency of 720 Hz with ( T = 36 \, \text{N} )
The frequency ( f ) is also inversely proportional to the length ( L ):
[
f \propto \frac{1}{L}
]
To achieve ( f_2 = 720 \, \text{Hz} ), the length ( L_2 ) can be determined relative to the initial ( f_1 = 288 \, \text{Hz} ) and ( L_1 = L ):
[
\frac{f_2}{f_1} = \frac{L_1}{L_2}
]
[
\frac{720}{288} = \frac{L_1}{L_2}
]
[
\frac{5}{2} = \frac{L_1}{L_2}
]
[
L_2 = \frac{2}{5} L_1
]
If ( L_1 = 1 \, \text{unit} ), then ( L_2 = 0.4 \, \text{units} ).
Answer: Reduce the length to 40% of its original value.
(iii) Fundamental frequency for given parameters
The fundamental frequency is given by:
[
f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}
]
Where:
- ( L = 0.5 \, \text{m} )
- ( T = 225 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 2205 \, \text{N} )
- Linear mass density ( \mu = \frac{\text{mass}}{\text{length}} = \frac{0.004}{0.5} = 0.008 \, \text{kg/m} )
Substitute values:
[
f = \frac{1}{2 \times 0.5} \sqrt{\frac{2205}{0.008}}
]
[
f = \frac{1}{1} \sqrt{\frac{2205}{0.008}}
]
[
f = \sqrt{275625} = 525 \, \text{Hz}
]
Answer: Fundamental frequency is ( 525 \, \text{Hz} ).
Explanation (300 Words)
The fundamental frequency of a vibrating wire depends on its tension, length, and mass per unit length. For a given wire under a specific tension, increasing the tension raises the frequency because the wave speed increases. The frequency is proportional to the square root of the tension. Therefore, doubling the tension increases the frequency by a factor of ( \sqrt{2} ), and tripling it increases it by ( \sqrt{3} ).
For part (i), the frequency change from 288 Hz to 432 Hz corresponds to a ( 3:2 ) ratio. Using the tension-frequency relationship, we find that increasing the tension to 81 N achieves this frequency.
For part (ii), the frequency also depends inversely on the string length. Shortening the length increases the frequency, which is why reducing the length to 40% of the original allows the wire to emit a note of 720 Hz while maintaining constant tension.
In part (iii), the formula for the fundamental frequency uses tension, string length, and mass per unit length. By substituting the given values, the frequency is calculated as 525 Hz. This shows how physical properties directly influence a string’s vibrational characteristics, crucial for musical instruments and engineering applications.