Answer :
The tension in the violin string needs to increase by approximately 5.9% to stay in tune with the organ pipe.
Certainly! To solve the problem, we first establish the relationship between the frequencies of the violin string and the organ pipe. At 0°C, both are in tune, hence their frequencies are equal.
For the violin string:
[tex]\[ f_{\text{violin}} = \frac{1}{2L_{\text{violin}}} \sqrt{\frac{T_{\text{violin}}}{\mu_{\text{violin}}}} \][/tex]
For the closed-open organ pipe:
[tex]\[ f_{\text{pipe}} = \frac{v_{0\°C}}{2L_{\text{organ}}} \][/tex]
where [tex]\( L_{\text{violin}} \) and \( \mu_{\text{violin}} \)[/tex] are the length and linear density of the violin string respectively, [tex]\( T_{\text{violin}} \)[/tex] is the tension in the violin string, [tex]\( v_{0\°C} \)[/tex] is the speed of sound at 0°C, and [tex]\( L_{\text{organ}} \)[/tex] is the length of the organ pipe.
Since the dimensions of the violin string remain constant, any change in frequency is solely due to the change in tension. So, we need the ratio of tensions at 0°C and 20°C:
[tex]\[ \frac{T_{20\°C}}{T_{\text{violin}}} = \frac{v_{0\°C}^2}{v_{20\°C}^2} \][/tex]
where [tex]\( v_{20\°C} \)[/tex] is the speed of sound at 20°C.
Now, let's calculate [tex]\( v_{0\°C} \)[/tex] and [tex]\( v_{20\°C} \)[/tex]:
[tex]\[ v_{0\°C} = \sqrt{\frac{\gamma R \cdot (0 + 273)}{M}} \][/tex]
[tex]\[ v_{20\°C} = \sqrt{\frac{\gamma R \cdot (20 + 273)}{M}} \][/tex]
After calculation, we find [tex]\( v_{0\°C} \approx 352.70 \text{ m/s} \) and \( v_{20\°C} \approx 342.78 \text{ m/s} \)[/tex].
Substituting these values into the tension ratio equation, we get [tex]\( \frac{T_{20\°C}}{T_{\text{violin}}} \approx 1.059 \)[/tex].
