How to calculate vibration in horizonatal string with a mass?
Answers
Answer:
Vibration, standing waves in a string. The fundamental and the first 5 overtones in the harmonic series.
A vibration in a string is a wave. Resonance causes a vibrating string to produce a sound with constant frequency, i.e. constant pitch. If the length or tension of the string is correctly adjusted, the sound produced is a musical tone. Vibrating strings are the basis of string instruments such as guitars, cellos, and pianos.
Wave Edit
The velocity of propagation of a wave in a string ( {\displaystyle v} v) is proportional to the square root of the force of tension of the string ( {\displaystyle T} T) and inversely proportional to the square root of the linear density ( {\displaystyle \rho } \rho ) of the string:
{\displaystyle v={\sqrt {T \over \rho }}.} {\displaystyle v={\sqrt {T \over \rho }}.}
This relationship was discovered by Vincenzo Galilei in the late 1500s.[citation needed]
Derivation Edit
Illustration for a vibrating string
Source:[1]
Let {\displaystyle \Delta x} \Delta x be the length of a piece of string, {\displaystyle m} m its mass, and {\displaystyle \rho } \rho its linear density. If the horizontal component of tension in the string is a constant, {\displaystyle T} T, then the tension acting on each side of the string segment is given by
{\displaystyle T_{1x}=T_{1}\cos(\alpha )\approx T.} T_{{1x}}=T_{1}\cos(\alpha )\approx T.
{\displaystyle T_{2x}=T_{2}\cos(\beta )\approx T.} T_{{2x}}=T_{2}\cos(\beta )\approx T.
If both angles are small, then the tensions on either side are equal and the net horizontal force is zero. From Newton's second law for the vertical component, the mass of this piece times its acceleration, {\displaystyle a} a, will be equal to the net force on the piece:
{\displaystyle \Sigma F_{y}=T_{1y}-T_{2y}=-T_{2}\sin(\beta )+T_{1}\sin(\alpha )=\Delta ma\approx \rho \Delta x{\frac {\partial ^{2}y}{\partial t^{2}}}.} {\displaystyle \Sigma F_{y}=T_{1y}-T_{2y}=-T_{2}\sin(\beta )+T_{1}\sin(\alpha )=\Delta ma\approx \rho \Delta x{\frac {\partial ^{2}y}{\partial t^{2}}}.}
Dividing this expression by {\displaystyle T} T and substituting the first and second equations obtains
{\displaystyle {\frac {\rho \Delta x}{T}}{\frac {\partial ^{2}y}{\partial t^{2}}}=-{\frac {T_{2}\sin(\beta )}{T_{2}\cos(\beta )}}+{\frac {T_{1}\sin(\alpha )}{T_{1}\cos(\alpha )}}=-\tan(\beta )+\tan(\alpha )} {\displaystyle {\frac {\rho \Delta x}{T}}{\frac {\partial ^{2}y}{\partial t^{2}}}=-{\frac {T_{2}\sin(\beta )}{T_{2}\cos(\beta )}}+{\frac {T_{1}\sin(\alpha )}{T_{1}\cos(\alpha )}}=-\tan(\beta )+\tan(\alpha )}
The tangents of the angles at the ends of the string piece are equal to the slopes at the ends, with an additional minus sign due to the definition of alpha and beta. Using this fact and rearranging provides
{\displaystyle {\frac {1}{\Delta x}}\left(\left.{\frac {\partial y}{\partial x}}\right|^{x+\Delta x}-\left.{\frac {\partial y}{\partial x}}\right|^{x}\right)={\frac {\rho }{T}}{\frac {\partial ^{2}y}{\partial t^{2}}}} {\displaystyle {\frac {1}{\Delta x}}\left(\left.{\frac {\partial y}{\partial x}}\right|^{x+\Delta x}-\left.{\frac {\partial y}{\partial x}}\right|^{x}\right)={\frac {\rho }{T}}{\frac {\partial ^{2}y}{\partial t^{2}}}}
In the limit that {\displaystyle \Delta x} \Delta x approaches zero, the left hand side is the definition of the second derivative of {\displaystyle y} y:
{\displaystyle {\frac {\partial ^{2}y}{\partial x^{2}}}={\frac {\rho }{T}}{\frac {\partial ^{2}y}{\partial t^{2}}}.} {\displaystyle {\frac {\partial ^{2}y}{\partial x^{2}}}={\frac {\rho }{T}}{\frac {\partial ^{2}y}{\partial t^{2}}}.}
This is the wave equation for {\displaystyle y(x,t)} y(x,t), and the coefficient of the second time derivative term is equal to {\displaystyle v^{-2}} v^{{-2}}; thus
{\displaystyle v={\sqrt {T \over \rho }},} {\displaystyle v={\sqrt {T \over \rho }},}
where {\displaystyle v} v is the speed of propagation of the wave in the string. (See the article on the wave equation for more about this). However, this derivation is only valid for vibrations of small amplitude; for those of large amplitude, {\displaystyle \Delta x} \Delta x is not a good approximation for the length of the string piece, the horizontal component of tension is not necessarily constant, and the horizontal tensions are not well approximated by. {\displaystyle T} T.