A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and time period is T. At the instance when its speed is half the maximum speed, its displacement x is `sqrt3/2 A`. Explanation: Express the relation for the velocity of a particle executing S.H.M. `v = omegasqrt(A^2 - x^2)` ......(1) Here, v is the velocity, `omega` is the angular speed, A is the amplitude and x is the displacement of the particle. Let `v_max` be the maximum speed of the particle. Given, `v = v_max/2` .....(2) Substitute (2) in (1), `v_max/2 = omegasqrt(A^2 - x^2)` .....(3) At the mean position, x = 0, the velocity is maximum. Therefore, `v_max/2 = omegasqrt(A^2 - x^2)` .....(4) = `omega sqrt(A^2 - 0^2)` = `omega A` Substitute (4) in (3) `omegaA/2 = omegasqrt(A^2 - x^2)` `A^2/4 = A^2 - x^2` `x^2 = A^2 = A^2/4 = (3A^2)/4` `x = sqrt(3)/2 A` Hence, the displacement of the particle executing S.H.M is x = `sqrt(3)/2 A`.
Option 4 : Alternately accelerated and retarded 
80 Questions 80 Marks 50 Mins
Concept:
\(u = {\rm{\omega \;}}\sqrt {{{\rm{a}}^2} - {y^2}} \) Where u = velocity, ω = angular velocity, a = amplitude and y = displacement. If y = 0 then; u = a ω, u is maximum when the particle is at its mean position of oscillation. If y = a then; u = 0. u is minimum when the particle is at any end position of the oscillation. Explanation:
So, acceleration and retardation alternatively present in SHM. India’s #1 Learning Platform Start Complete Exam Preparation
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Mock Tests & Quizzes Trusted by 3.3 Crore+ Students Answer Hint The maximum velocity of a particle in simple harmonic motion is given by $v = r\omega $ and the maximum acceleration of a particle in simple harmonic motion is given by $a = r{\omega ^2}$ . Divide one by the other and find the value $\omega $ . Use this value in either of the equations to find the value of the amplitude of the particle in simple harmonic motion. Complete Step by step answer Let the maximum velocity of the particle in simple harmonic motion be $v$ . Let the maximum acceleration of the particle be $a$ . Let the amplitude of the particle be $r$ . Let $\omega $ be the frequency of the particle executing simple harmonic motion. It is given in the question that the maximum velocity of the particle is $20cm/s$ and the maximum acceleration of the particle is $80cm/{s^2}$ . The maximum velocity of a particle executing simple harmonic motion is given by $v = r\omega $ Substituting the value of maximum velocity, we get$20 = r\omega $ The maximum acceleration of a particle executing simple harmonic motion is given by $a = r{\omega ^2}$ Substituting the maximum value of acceleration, we get$80 = r{\omega ^2}$ Dividing the above equation for maximum acceleration by the already written equation for maximum velocity, we get$\dfrac{{80}}{{20}} = \omega $ $ \Rightarrow \omega = 4Hz$ Substituting this value of frequency into the equation for maximum velocity, we get$20 = r \times 4$ By cross multiplication, we get$r = \dfrac{{20}}{4}$ $ \Rightarrow r = 5cm$ That is, the maximum amplitude calculated is $5cm$ .Hence, option (C) is the correct option. Note Simple harmonic motion is a type of periodic motion. In simple harmonic motion, the amplitude of a particle will also be its maximum displacement from the mean position of the particle. The particle will have maximum velocity when it is at the mean position and the particle will have maximum acceleration when it is at its extreme positions. |
When a particle performing SHM is at mean position it has?
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