Physics - Sound Wave Question with Solution | TestHub

PhysicsSound WaveStanding waveMedium2 minPYQ_2020

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A string of length $1.5 m$ with its two ends clamped is vibrating in the fundamental mode. The amplitude at the centre of the string is $4 mm$ . The minimum distance between the two points having amplitude of $2 mm$ is:

Options:

Answer:

Solution:

$λ = 2 l = 3 m$

Equation of standing wave

$y = 2 A \sin k x \cos ω t$

$y = A$ as amplitude is 2A.

$A = 2 A \sin k x$

$\frac{2 π}{λ} x_{1} = \frac{π}{6} \Rightarrow x_{1} = \frac{1}{4} m$

And $\frac{2 π}{λ} . x_{2} = \frac{5 π}{6} \Rightarrow x_{2} = 1.25 m \Rightarrow x_{2} - x_{1} = 1 m$

Stream:JEESubject:PhysicsTopic:Sound WaveSubtopic:Standing wave

⏱ 2mℹ️ Source: PYQ_2020

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