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For the rigid rod shown in diagram, density varies according to $ρ = 1 + x$ with distance $x$ from left end. If amplitude of vibrator performing SHM is A, amplitude of point at a distance x = 1 is

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$0.707 A$

$A$

$0.414 A$

$0.577 A$

answer is A.

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Detailed Solution

Energy density of wave is given by
$\frac{1}{2} ρ A^{2} ω^{2}$
Equating energy densities at x = 0 and x = 1, we get
$\frac{1}{2} (1 + 0) A^{2} = \frac{1}{2} (1 + 1) A^{'^{2}} \Rightarrow A^{'} = \frac{A}{\sqrt{2}} = 0.707 A$

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