A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, $y (x, t) = (0.01 m) \sin [(62.8 m^{- 1}) \times] \cos [(628 s^{- 1}) t]$ . Assuming $π = 3.14$ , the correct statement(s) is are

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The number of nodes is 5.

The length of the string is 0.25m.

The maximum displacement of the midpoint of the string its equilibrium position is 0.01m.

The fundamental frequency is 100 Hz.

answer is B, C.

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Detailed Solution

(A) The are 5 complete loops.

Total number of nodes = 6

(B) $ω = 628 \sec^{- 1}$

$k = 62.8 m^{- 1} = \frac{2 π}{λ} \Rightarrow λ = \frac{1}{10}$

$υ_{w} = \frac{ω}{k} = \frac{628}{62.8} = 10 m s^{- 1}$

$L = \frac{5 λ}{2} = 0.25$

(D) $f = \frac{υ}{2 l} = \frac{10}{2 \times 0.25} = 20 H z .$

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