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Standard XII

Physics

Springs in Parallel and Series

A particle mo...

Question

A particle moves under the force $F = (x^{2} - 6 x) N$ , where $x$ is in metres. For small displacements from the origin, what is the force constant in $N/m$ in the simple harmonic motion approximation?

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Solution

$F = x^{2} - 6 x$
For small displacement from the origin, the force is
$F^{'} = F (x + Δ x) - F (x) = (x + Δ x)^{2} - 6 (x + Δ x) - x^{2} + 6 x = 2 x Δ x - 6 Δ x$
$= - (6 - 2 x) Δ x$
For $x = 0$
$F^{'} = - 6 Δ x = - k Δ x \Rightarrow k = 6 N/m$

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A particle moves under the force F=(x2−6x) N, where x is in metres. For small displacements from the origin, what is the force constant in N/m in the simple harmonic motion approximation?

F=x2−6x For small displacement from the origin, the force is F′=F(x+Δx)−F(x)=(x+Δx)2−6(x+Δx)−x2+6x=2xΔx−6Δx =−(6−2x)Δx For x=0 F′=−6Δx=−kΔx⇒k=6 N/m

A particle moves under the force $F = (x^{2} - 6 x) N$ , where $x$ is in metres. For small displacements from the origin, what is the force constant in $N/m$ in the simple harmonic motion approximation?

$F = x^{2} - 6 x$
For small displacement from the origin, the force is
$F^{'} = F (x + Δ x) - F (x) = (x + Δ x)^{2} - 6 (x + Δ x) - x^{2} + 6 x = 2 x Δ x - 6 Δ x$
$= - (6 - 2 x) Δ x$
For $x = 0$
$F^{'} = - 6 Δ x = - k Δ x \Rightarrow k = 6 N/m$