A uniform string fixed at both ends is vibrating in 3rd harmonic and equation y - 4 cm sin- 0-8 cm - 1 - cos- 400 - s - 1 t -The length of the vibrating string is

The correct option is C 11.8 m y=4sin #160; ( 0.8x ) cos #160; ( 400π t ) #160; w=400π =2π f f=200 Hz v= w k = 400π× 100 ...

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Standard XII

Physics

Wave Speed expression

A uniform str...

Question

A uniform string fixed at both ends is vibrating in 3rd harmonic and equation $y = 4 (c m)$ $sin [(0.8 {c m}^{- 1}) \times] cos [(400 π s^{- 1}) t]$ The length of the vibrating string is

6.75 m

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12.45 m

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11.8 m

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18.7 m

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Solution

The correct option is C $11.8 m$
$y = 4 sin (0.8 x) cos (400 π t) w = 400 π = 2 π f f = 200 H z v = \frac{w}{k} = \frac{400 π \times 100}{0.8} m / s 3 \cdot \frac{v}{2 l} = f \Rightarrow 200 = \frac{3 \times 400 π \times 100}{2 \times l \times 0.8} \Rightarrow l = \frac{3 \times 400 π}{4 \times 0.8} = \frac{300 π}{0.8} c m = 11.8 m$
Hence,
option $(C)$ is correct answer.

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A uniform string fixed at both ends is vibrating in 3rd harmonic and equation y=4(cm) sin[(0.8cm−1)×]cos[(400πs−1)t]The length of the vibrating string is

The correct option is C 11.8my=4sin(0.8x)cos(400πt)w=400π=2πff=200Hzv=wk=400π×1000.8m/s3⋅v2l=f⇒200=3×400π×1002×l×0.8⇒l=3×400π4×0.8=300π0.8cm=11.8mHence,option (C) is correct answer.

A uniform string fixed at both ends is vibrating in 3rd harmonic and equation $y = 4 (c m)$ $sin [(0.8 {c m}^{- 1}) \times] cos [(400 π s^{- 1}) t]$ The length of the vibrating string is