Question

dx 9. x= A cos (nt + a), JEI N STER 791 A, QWERTGRI , (simple harmonic motion) का अवकल समीकरण बनाएँ। [Form the differential equation of simple harmonic motion given by x= A cos (nt + a), where n is fixed and A, a are parameters.)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\:x = A \: cos(nt + a)$

On differentiating both sides w. r. t. t, we get

$\rm :\longmapsto\:\dfrac{d}{dt}x =\dfrac{d}{dt} A \: cos(nt + a)$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}cosx \: = \: - \: sinx}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dx}{dt} = - A \: sin(nt + a)\dfrac{d}{dt}(nt + a)$

$\rm :\longmapsto\:\dfrac{dx}{dt} = - A \: sin(nt + a)(n \times 1 + 0)$

$\rm :\longmapsto\:\dfrac{dx}{dt} = - An \: sin(nt + a)$

On differentiating both sides w. r. t. t, we get

$\rm :\longmapsto\:\dfrac{d}{dt}\dfrac{dx}{dt} = \dfrac{d}{dt}[- An \: sin(nt + a)]$

$\rm :\longmapsto\:\dfrac{ {d}^{2}x }{ {dt}^{2} } = - An \: \dfrac{d}{dt}sin(nt + a)$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}sinx \: = \: cosx}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2}x }{ {dt}^{2} } = - An \: cos(nt + a) \: \dfrac{d}{dt}(nt + a)$

$\rm :\longmapsto\:\dfrac{ {d}^{2}x }{ {dt}^{2} } = -nx \:(n \times 1 + 0) \: \: \: \: \{ \because \: x = Acos(nt + a) \}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}x }{ {dt}^{2} } = -nx \times n$

$\rm :\longmapsto\:\dfrac{ {d}^{2}x }{ {dt}^{2} } = - {n}^{2} x$

$\rm\implies \:\boxed{\tt{ \: \: \dfrac{ {d}^{2}x }{ {dt}^{2} } + {n}^{2} x = 0 \: \: }}$

is the required differential equation.

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MORE TO KNOW

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\:x = A \: cos(nt + a)$

On differentiating both sides w. r. t. t, we get

$\rm :\longmapsto\:\dfrac{d}{dt}x =\dfrac{d}{dt} A \: cos(nt + a)$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}cosx \: = \: - \: sinx}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dx}{dt} = - A \: sin(nt + a)\dfrac{d}{dt}(nt + a)$

$\rm :\longmapsto\:\dfrac{dx}{dt} = - A \: sin(nt + a)(n \times 1 + 0)$

$\rm :\longmapsto\:\dfrac{dx}{dt} = - An \: sin(nt + a)$

On differentiating both sides w. r. t. t, we get

$\rm :\longmapsto\:\dfrac{d}{dt}\dfrac{dx}{dt} = \dfrac{d}{dt}[- An \: sin(nt + a)]$

$\rm :\longmapsto\:\dfrac{ {d}^{2}x }{ {dt}^{2} } = - An \: \dfrac{d}{dt}sin(nt + a)$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}sinx \: = \: cosx}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2}x }{ {dt}^{2} } = - An \: cos(nt + a) \: \dfrac{d}{dt}(nt + a)$

$\rm :\longmapsto\:\dfrac{ {d}^{2}x }{ {dt}^{2} } = -nx \:(n \times 1 + 0) \: \: \: \: \{ \because \: x = Acos(nt + a) \}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}x }{ {dt}^{2} } = -nx \times n$

$\rm :\longmapsto\:\dfrac{ {d}^{2}x }{ {dt}^{2} } = - {n}^{2} x$

$\rm\implies \:\boxed{\tt{ \: \: \dfrac{ {d}^{2}x }{ {dt}^{2} } + {n}^{2} x = 0 \: \: }}$

is the required differential equation.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$