The phase difference between two points is π 3 . If the frequency of waves is 50 Hz , then, the distance between two points is (Given v = 330 m s - 1 )

From the relation, ∆ ϕ = 2 π λ × ∆ x ⟹ ∆ x = λ 2 π × ∆ ϕ … i Also, λ = v n … i i Now, from Eqs.(i) and (ii), we get ∆ x = v 2 π n × ∆ ϕ ⟹ ∆ x = 330 2 π × 50 × π 3 or ∆ x = 1 . 1 m

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Physics - Wave on String Question with Solution | TestHub

PhysicsWave on StringEquation/graphEasy2 minPYQ_2020

PhysicsEasysingle choice

The phase difference between two points is $\frac{π}{3}$ . If the frequency of waves is $50 Hz$ , then, the distance between two points is (Given $v = 330 m s^{- 1}$ )

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Solution:

From the relation,

$∆ ϕ = \frac{2 π}{λ} \times ∆ x$

$⟹ ∆ x = \frac{λ}{2 π} \times ∆ ϕ \dots (i)$

Also, $λ = \frac{v}{n} \dots (i i)$

Now, from Eqs.(i) and (ii), we get

$∆ x = \frac{v}{2 π n} \times ∆ ϕ$

$⟹ ∆ x = \frac{330}{2 π \times 50} \times \frac{π}{3}$

$or ∆ x = 1.1 m$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Wave on StringSubtopic:Equation/graph

⏱ 2mℹ️ Source: PYQ_2020

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