A vibrating string of length 80 cm is vibrating with a fundamental frequency of 400 Hz. In the fundamental mode, the maximum displacement at the middle is 2cm. The tension in the string is 10N. Assume that tension in the string to be constant as oscillation is very small. The maximum value of the component of force on the end support, perpendicular to the equilibrium position of the string is

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1.250 N

0.4750 N

1.875 N

0.785 N

answer is D.

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Detailed Solution

$F_{y} = T \sin θ = T \tan θ F_{y} = T \sin θ = T \tan θ = T (\frac{\partial y}{\partial x}) y = A \sin [ω (t - \frac{x}{V})] ω = \frac{2 π f}{λ} = \frac{2 π f}{2 l} = \frac{π f}{80}$
A = 2cm
$y = 2 \sin [ω t - \frac{π x}{80}] c m \frac{\partial y}{\partial x} = 2 \cos [ω t = \frac{π x}{80}] \times (\frac{π}{80}) = - \frac{2 π}{80} \cos [ω t - \frac{π x}{80}] = - \frac{π}{40} \cos [ω t - \frac{π t}{80}] = F_{y} = T \frac{\partial y}{\partial x} 10 = - \frac{π}{40} \cos (ω t - \frac{π x}{80})$