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A body of mass 600 g is attached to a spring of spring constant k = 100 N/m and it is performing damped oscillations. If damping constant is 0.2 and driving force is $F = F_{0} \cos (ωt),$ where $F_{0} = 20 N$ . Find the amplitude of oscillation at resonance :

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a

4.1 m

b

0.57 m

c

7.7 m

d

0.98 m

answer is C.

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Detailed Solution

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For damped oscillations,
$A = \frac{F_{0} / m}{\sqrt{{(ω^{2} - ω_{0}^{2})}^{2} + {(bω / m)}^{2}}}$
At resonance, $ω = ω_{0} so A = \frac{F_{0}}{bω}$
Here,
$ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{0.6}} = 13 rad / s So, A = \frac{20}{(0.2) 13} = 7.7 m$

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A body of mass 600 g is attached to a spring of spring constant k = 100 N/m and it is performing damped oscillations. If damping constant is 0.2 and driving force is F=F0 cos ωt,where F0=20N. Find the amplitude of oscillation at resonance :