The figure shows a snap shot of a vibrating string at t -0- The particle P is observed moving up with velocity 20 -3 cm - s- The tangent at P makes an angle 60- with x-axis-The speed and the direction of the wave isA- 40 cm - s along positive x-axisB- 20 cm - s along negative x-axisC- 30 cm - s along positive x-axisD- 60 cm - s along negative x-axis

The correct option is B 20 cm/s along negative x axis dy/dx=tan 60^∘= √(3), v_p= v dy/dx⇒ 20 √(3)= v √(3) ⇒ v=20 cm/s ...

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Byju's Answer

Standard XII

Physics

Wave in Negative X-Direction

The figure sh...

Question

The figure shows a snap shot of a vibrating string at t = 0. The particle P is observed moving up with velocity $20 \sqrt{3}$ cm/s. The tangent at P makes an angle $60^{\circ}$ with x-axis.

The speed and the direction of the wave is

30 cm/s along positive x-axis

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60 cm/s along negative x-axis

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40 cm/s along positive x-axis

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20 cm/s along negative x-axis

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Solution

The correct option is D
20 cm/s along negative x-axis

$\frac{d y}{d x} = t a n 60^{\circ} = \sqrt{3}, v_{p} = - v \frac{d y}{d x} \Rightarrow 20 \sqrt{3} = - v \sqrt{3}$
$\Rightarrow$ v=20 cm/s

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The figure shows a snap shot of a vibrating string at t = 0. The particle P is observed moving up with velocity 20√3cm/s. The tangent at P makes an angle 60∘ with x-axis. The speed and the direction of the wave is

The correct option is D 20 cm/s along negative x-axis dydx=tan60∘=√3, vp=−vdydx⇒20√3=−v√3 ⇒ v=20 cm/s

The figure shows a snap shot of a vibrating string at t = 0. The particle P is observed moving up with velocity $20 \sqrt{3}$ cm/s. The tangent at P makes an angle $60^{\circ}$ with x-axis.

The speed and the direction of the wave is

The correct option is D
20 cm/s along negative x-axis

$\frac{d y}{d x} = t a n 60^{\circ} = \sqrt{3}, v_{p} = - v \frac{d y}{d x} \Rightarrow 20 \sqrt{3} = - v \sqrt{3}$
$\Rightarrow$ v=20 cm/s