A particle of mass m connected with a spring of force constant k is executing simple harmonic motion with amplitude A . When it is at + A 2 from equilibrium position moving towards right, it receives an impulse which doubles its velocity without changing direction. velocity of the particle after the impulse,

Time period of oscillation of the particle ω = k m or ω = k m … i We define velocity of the particle in SHM by relation v = ω A 2 - y 2 = 2 π T A 2 - y 2 … ii Here y = A 2 , hence the velocity of the particle at this position, v = 2 π T A 2 - A 2 2 = π 3 A T After impulse the velocity of the particle doubles, hence velocity of the particle after impulse v ' = 2 v = 2 π 3 A T … iii After impulse the time period of the particle and angular frequency remains unchanged as neither mass nor spring constant is changing, but amplitude of oscillation will change. Let new amplitude is A 1 , Then again using Eq. (ii), we get v ' = 2 π 3 T = 2 π T A 1 2 - y 2 3 A = A 1 2 - A 2 2 ⇒ A 1 = 13 2 A For calculating time taken by the block to reach the right extreme position, let us draw phasor circle correspond to this situation. The circle corresponding to radius A represents the situation just before impulse and the circle of radius A represents phasor circle after the impulse. As the direction of motion of the particle remains unchanged, hence we need to find the time in reaching from position P 1 to Q or time corresponding to change in phase angle θ on the phasor circle corresponding to radius A 1 . Hence cos θ = A / 2 A 1 = A · 2 2 13 A = 1 13 But θ = ω t ⇒ cos - 1 1 13 = ω t cos - 1 1 13 = k m t ⇒ t = m k cos - 1 1 13

A particle of mass m connected with a spring of force constant k is executing simple harmonic motion with amplitude A . When it is at + A 2 from equilibrium position moving towards right, it receives an impulse which doubles its velocity without changing direction. amplitude after the impulse, and

Time period of oscillation of the particle, ω = k m … i We define velocity of the particle in SHM by relation v = ω A 2 - y 2 = 2 π T A 2 - y 2 … ii Here y = A 2 , hence the velocity of the particle at this position, v = 2 π T A 2 - A 2 2 = π 3 A T After impulse the velocity of the particle doubles, hence velocity of the particle after impulse v ' = 2 v = 2 π 3 A T … iii After impulse the time period of the particle and angular frequency remains unchanged as neither mass nor spring constant is changing but amplitude of oscillation will change. Let new amplitude is A 1 , then again using Eq. (ii), we get v ' = 2 π 3 T = 2 π T A 2 - y 2 3 A = A 1 2 - A 2 2 ⇒ A 1 = 13 2 A For calculating time taken by the block to reach the right extreme position, let us draw phasor circle correspond to this situation. The circle corresponding to radius A represents the situation just before impulse and the circle of radius A represents phasor circle after the impulse. As the direction of motion of the particle remains unchanged, hence we need to find the time in reaching from position P 1 to Q or time corresponding to change in phase angle θ on the phasor circle corresponding to radius A 1 . Hence cos θ = A / 2 A 1 = A × 2 2 13 A = 1 13 But θ = ω t ⇒ cos - 1 1 13 = ω t cos - 1 1 13 = k m t ⇒ t = m k cos - 1 1 13

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A spring of stiffness constant $k$ and natural length $l$ is cut into two parts of length $3 l / 4$ and $l / 4,$ respectively, and an arrangement is made as shown in Figs. (a) and (b). If the mass is slightly displaced, find the time period of oscillation.

Important Questions on Linear and Angular Simple Harmonic Motion

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PHYSICS FOR JOINT ENTRANCE EXAMINATION WAVES AND THERMODYNAMICS>Chapter 5 - Linear and Angular Simple Harmonic Motion>ILLUSTRATIONS>Q 5.27

A block of mass $m$ hangs from a smooth light pulley and by an inextensible string fitted a spring of stiffness $k$ as shown in case and case

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PHYSICS FOR JOINT ENTRANCE EXAMINATION WAVES AND THERMODYNAMICS>Chapter 5 - Linear and Angular Simple Harmonic Motion>ILLUSTRATIONS>Q 5.27

A block of mass $m$ hangs from a smooth light pulley and by an inextensible string fitted a spring of stiffness $k$ as shown in case and case

Find the period of oscillation of the block.

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PHYSICS FOR JOINT ENTRANCE EXAMINATION WAVES AND THERMODYNAMICS>Chapter 5 - Linear and Angular Simple Harmonic Motion>ILLUSTRATIONS>Q 5.28

Two light springs of force constants $k_{1}$ and $k_{2}$ and a block of mass $m$ are in one line $A B$ on a smooth horizontal table such that one end of each spring is fixed to rigid supports and the other end is free as shown in the figure. The distance $C D$ between the free ends of the spring is $60 cm$ . If the block moves along $A B$ with a velocity $120 cm s^{- 1}$ in between the springs, calculate the period of oscillation of the block $(k_{1} = 1.8 N m^{- 1}, k_{2} = 3.2 N m^{- 1}, m = 200 g)$ .

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PHYSICS FOR JOINT ENTRANCE EXAMINATION WAVES AND THERMODYNAMICS>Chapter 5 - Linear and Angular Simple Harmonic Motion>ILLUSTRATIONS>Q 5.29

A particle of mass $m$ connected with a spring of force constant $k$ is executing simple harmonic motion with amplitude $A .$ When it is at $+ \frac{A}{2}$ from equilibrium position moving towards right, it receives an impulse which doubles its velocity without changing direction.

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velocity of the particle after the impulse,

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PHYSICS FOR JOINT ENTRANCE EXAMINATION WAVES AND THERMODYNAMICS>Chapter 5 - Linear and Angular Simple Harmonic Motion>ILLUSTRATIONS>Q 5.29

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amplitude after the impulse, and

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PHYSICS FOR JOINT ENTRANCE EXAMINATION WAVES AND THERMODYNAMICS>Chapter 5 - Linear and Angular Simple Harmonic Motion>ILLUSTRATIONS>Q 5.29

A particle of mass $m$ connected with a spring of force constant $k$ is executing simple harmonic motion with amplitude $A .$ When it is at $+ \frac{A}{2}$ from equilibrium position moving towards the right, it receives an impulse which doubles its velocity without changing direction.

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time taken by the block to reach the right extreme position.

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PHYSICS FOR JOINT ENTRANCE EXAMINATION WAVES AND THERMODYNAMICS>Chapter 5 - Linear and Angular Simple Harmonic Motion>ILLUSTRATIONS>Q 5.30

A block of mass $m$ connected to a spring of force constant $k$ is performing simple harmonic motion. Initially, the block is compressed by a distance $A$ and released. An elastic wall is located in front of block at a distance of $A / 2 .$ Find the time period of oscillation of this block.

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PHYSICS FOR JOINT ENTRANCE EXAMINATION WAVES AND THERMODYNAMICS>Chapter 5 - Linear and Angular Simple Harmonic Motion>ILLUSTRATIONS>Q 5.31

Figures (a) represent spring-block system. If $m$ is displaced slightly, find the time period of oscillation of the system.

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Both the cases are as follows:

A spring of stiffness constant k and natural length l is cut into two parts of length 3l/4 and l/4, respectively, and an arrangement is made as shown in Figs. (a) and (b). If the mass is slightly displaced, find the time period of oscillation.

Important Questions on Linear and Angular Simple Harmonic Motion

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A spring of stiffness constant $k$ and natural length $l$ is cut into two parts of length $3 l / 4$ and $l / 4,$ respectively, and an arrangement is made as shown in Figs. (a) and (b). If the mass is slightly displaced, find the time period of oscillation.