Show that Simple Harmonic motion is Isochronous.
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Answer:
In Newtonian mechanics, for one-dimensional simple harmonic motion, the equation of motion, which is a second-order linear ordinary differential equation with constant coefficients, can be obtained by means of Newton's 2nd law and Hooke's law for a mass on a spring.
{\displaystyle F_{\mathrm {net} }=m{\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}=-kx,} {\displaystyle F_{\mathrm {net} }=m{\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}=-kx,}
where m is the inertial mass of the oscillating body, x is its displacement from the equilibrium (or mean) position, and k is a constant (the spring constant for a mass on a spring).
Therefore,
{\displaystyle {\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}=-{\frac {k}{m}}x,} {\displaystyle {\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}=-{\frac {k}{m}}x,}
Solving the differential equation above produces a solution that is a sinusoidal function.
{\displaystyle x(t)=c_{1}\cos \left(\omega t\right)+c_{2}\sin \left(\omega t\right)} {\displaystyle x(t)=c_{1}\cos \left(\omega t\right)+c_{2}\sin \left(\omega t\right)}
This equation can be written in the form:
{\displaystyle x(t)=A\cos \left(\omega t-\varphi \right),} {\displaystyle x(t)=A\cos \left(\omega t-\varphi \right),}
where
{\displaystyle \omega ={\sqrt {\frac {k}{m}}},\qquad A={\sqrt {{c_{1}}^{2}+{c_{2}}^{2}}},\qquad \tan \varphi ={\frac {c_{2}}{c_{1}}},} {\displaystyle \omega ={\sqrt {\frac {k}{m}}},\qquad A={\sqrt {{c_{1}}^{2}+{c_{2}}^{2}}},\qquad \tan \varphi ={\frac {c_{2}}{c_{1}}},}
In the solution, c1 and c2 are two constants determined by the initial conditions (specifically, the initial position at time t = 0 is c1, while the initial velocity is c2ω), and the origin is set to be the equilibrium position.[A] Each of these constants carries a physical meaning of the motion: A is the amplitude (maximum displacement from the equilibrium position), ω = 2πf is the angular frequency, and φ is the initial phase.[B]
Using the techniques of calculus, the velocity and acceleration as a function of time can be found:
{\displaystyle v(t)={\frac {\mathrm {d} x}{\mathrm {d} t}}=-A\omega \sin(\omega t-\varphi ),} v(t)={\frac {\mathrm {d} x}{\mathrm {d} t}}=-A\omega \sin(\omega t-\varphi ),
Speed:
{\displaystyle {\omega }{\sqrt {A^{2}-x^{2}}}} {\omega }{\sqrt {A^{2}-x^{2}}}
Maximum speed: v=ωA (at equilibrium point)
{\displaystyle a(t)={\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}=-A\omega ^{2}\cos(\omega t-\varphi ).} a(t)={\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}=-A\omega ^{2}\cos(\omega t-\varphi ).
Maximum acceleration: Aω2 (at extreme points)
By definition, if a mass m is under SHM its acceleration is directly proportional to displacement.
{\displaystyle a(x)=-\omega ^{2}x.} {\displaystyle a(x)=-\omega ^{2}x.}
where
{\displaystyle \omega ^{2}={\frac {k}{m}}} {\displaystyle \omega ^{2}={\frac {k}{m}}}
Since ω = 2πf,
{\displaystyle f={\frac {1}{2\pi }}{\sqrt {\frac {k}{m}}},} f={\frac {1}{2\pi }}{\sqrt {\frac {k}{m}}},
and, since T =
1
/
f
where T is the time period,
{\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}.} T=2\pi {\sqrt {\frac {m}{k}}}.
These equations demonstrate that the simple harmonic motion is isochronous (the period and frequency are independent of the amplitude and the initial phase of the motion).