
The amplitude of a particle executing S.H.M. is $ 4cm $ . At the mean position the speed of the particle is $ 16cm{s^{ - 1}} $ . The distance of the particle from the mean position at which the speed of the particle becomes $ 8\sqrt 3 cm{s^{ - 1}} $ , will be:
(A) $ 2\sqrt 3 cm $
(B) $ \sqrt 3 cm $
(C) $ 1cm $
(D) $ 2cm $
Answer
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Hint :Here, in the given question it is mentioned that we are talking about the S.H.M. so, we must know the properties and all the terms of simple harmonic oscillation such as amplitude, mean position, time period and speed of the particle in simple harmonic motion.
The useful formulas are: $ {V_{\max }} = A\omega $
$ V = \omega \sqrt {{A^2} - {y^2}} $
Complete Step By Step Answer:
let amplitude of S.H.M. be $ A $ , at mean position the velocity of the particle is maximum hence let velocity be $ {V_{\max }} $ , let the velocity of particle away from the mean position be $ V $ , let $ \omega $ be the angular velocity of the particle and $ y $ be the position of the particle away from mean position.
Given data: $ {V_{\max }} = 16cm{s^{ - 1}} $
$ A = 4cm $
$ V = 8\sqrt 3 cm{s^{ - 1}} $
Thus, let us use the formula of maximum velocity to find angular velocity such that
$ {V_{\max }} = A\omega $
$ \Rightarrow 16 = 4 \times \omega $
$ \Rightarrow \omega = 4 $
Now, we have velocity of particle away from the mean position such that
$ V = 8\sqrt 3 cm{s^{ - 1}} $
$ \Rightarrow V = \omega \sqrt {{A^2} - {y^2}} $
$ \Rightarrow 8\sqrt 3 = 4\sqrt {{4^2} - {y^2}} $
On squaring both the sides, we get
$ \Rightarrow {(8\sqrt 3 )^2} = {4^2}\left( {{4^2} - {y^2}} \right) $
$ \Rightarrow 8 \times 8 \times 3 = 4 \times 4 \times (16 - {y^2}) $
$ \Rightarrow 12 = 16 - {y^2} $
On solving the above equation, we get
$ \Rightarrow y = 2cm $
Thus the particle position away from mean position is $ 2cm $
Hence, the correct answer is option D.
Note :
We have to study the problem carefully and apply suitable formulae to solve this problem. We must know the conditions of having maximum velocity of the particle at which position and also apply suitable formula to find the position of the particle.
The useful formulas are: $ {V_{\max }} = A\omega $
$ V = \omega \sqrt {{A^2} - {y^2}} $
Complete Step By Step Answer:
let amplitude of S.H.M. be $ A $ , at mean position the velocity of the particle is maximum hence let velocity be $ {V_{\max }} $ , let the velocity of particle away from the mean position be $ V $ , let $ \omega $ be the angular velocity of the particle and $ y $ be the position of the particle away from mean position.
Given data: $ {V_{\max }} = 16cm{s^{ - 1}} $
$ A = 4cm $
$ V = 8\sqrt 3 cm{s^{ - 1}} $
Thus, let us use the formula of maximum velocity to find angular velocity such that
$ {V_{\max }} = A\omega $
$ \Rightarrow 16 = 4 \times \omega $
$ \Rightarrow \omega = 4 $
Now, we have velocity of particle away from the mean position such that
$ V = 8\sqrt 3 cm{s^{ - 1}} $
$ \Rightarrow V = \omega \sqrt {{A^2} - {y^2}} $
$ \Rightarrow 8\sqrt 3 = 4\sqrt {{4^2} - {y^2}} $
On squaring both the sides, we get
$ \Rightarrow {(8\sqrt 3 )^2} = {4^2}\left( {{4^2} - {y^2}} \right) $
$ \Rightarrow 8 \times 8 \times 3 = 4 \times 4 \times (16 - {y^2}) $
$ \Rightarrow 12 = 16 - {y^2} $
On solving the above equation, we get
$ \Rightarrow y = 2cm $
Thus the particle position away from mean position is $ 2cm $
Hence, the correct answer is option D.
Note :
We have to study the problem carefully and apply suitable formulae to solve this problem. We must know the conditions of having maximum velocity of the particle at which position and also apply suitable formula to find the position of the particle.
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