Respuesta :
Answer:
(i) [tex]\dfrac{df}{dL}=-\dfrac{1}{2L^2}\sqrt{\dfrac{T}{\rho}}[/tex]
(ii) [tex]\dfrac{df}{dT}=\dfrac{1}{4L\sqrt{T\rho}}[/tex]
(iii) [tex]\dfrac{df}{d\rho}=-\dfrac{\sqrt{T}}{4L\rho^{-\frac{3}{2}}}}[/tex]
Step-by-step explanation:
Let as consider the frequency (in Hz) of a vibrating violin string is given by
[tex]f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\rho}}[/tex]
(i)
Differentiate f with respect L (assuming T and rho are constants).
[tex]\dfrac{df}{dL}=\dfrac{d}{dL}\dfrac{1}{2L}\sqrt{\dfrac{T}{\rho}}[/tex]
Taking out constant terms.
[tex]\dfrac{df}{dL}=\dfrac{1}{2}\sqrt{\dfrac{T}{\rho}}\dfrac{d}{dL}\dfrac{1}{L}[/tex]
[tex]\dfrac{df}{dL}=\dfrac{1}{2}\sqrt{\dfrac{T}{\rho}}(-\dfrac{1}{L^2})[/tex]
[tex]\dfrac{df}{dL}=-\dfrac{1}{2L^2}\sqrt{\dfrac{T}{\rho}}[/tex]
(ii)
Differentiate f with respect T (assuming L and rho are constants).
[tex]\dfrac{df}{dT}=\dfrac{d}{dT}\dfrac{1}{2L}\sqrt{\dfrac{T}{\rho}}[/tex]
Taking out constant terms.
[tex]\dfrac{df}{dT}=\dfrac{1}{2L}\sqrt{\dfrac{1}{\rho}}\dfrac{d}{dT}\sqrt{T}}[/tex]
[tex]\dfrac{df}{dT}=\dfrac{1}{2L}\sqrt{\dfrac{1}{\rho}}(\dfrac{1}{2\sqrt{T}})[/tex]
[tex]\dfrac{df}{dT}=\dfrac{1}{4L\sqrt{T\rho}}[/tex]
(iii)
Differentiate f with respect rho (assuming L and T are constants).
[tex]\dfrac{df}{d\rho}=\dfrac{d}{d\rho}\dfrac{1}{2L}\sqrt{\dfrac{T}{\rho}}[/tex]
Taking out constant terms.
[tex]\dfrac{df}{d\rho}=\dfrac{\sqrt{T}}{2L}\dfrac{d}{d\rho}(\rho)^{-\frac{1}{2}}}[/tex]
[tex]\dfrac{df}{d\rho}=\dfrac{\sqrt{T}}{2L}(-\dfrac{1}{2}(\rho)^{-\frac{3}{2}}})[/tex]
[tex]\dfrac{df}{d\rho}=-\dfrac{\sqrt{T}}{4L\rho^{-\frac{3}{2}}}}[/tex]
