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Ch08 Periodic Motion

Lecture slides and notes for Ch08 Periodic Motion from the Physics Homework Notes module in HSC Resources by Md Ahbab. 14 pages.

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PHYSICS 1ST PAPER CHAPTER 8 Periodic Motion Question 1 d2π‘₯ d𝑑2 + πœ”2π‘₯= 0 is a differential equation of a simple harmonic motion. When asked to write its solution, Shamim wrote π‘₯= 𝐡cos(πœ”π‘‘+ 𝛼) and Rimjhim wrote π‘₯= 𝐢tan(πœ”π‘‘+ πœ‘). (a) What is called periodic motion? (b) What do you mean by temporal periodicity? (c) Show that π‘₯= 𝐴sin(πœ”π‘‘+ 𝛿) is a solution of the differential equation of simple harmonic motion. (d) Verify by mathematical analysis whether the solutions of Shamim and Rimjhim are proper or not. Solution (a) A motion that repeats itself after equal intervals of time is called periodic motion. (b) Some events repeat after a fixed interval of time. This repetition with time is called temporal periodicity. The fixed interval is the time period 𝑇. The state of the body at a time 𝑑returns at the time 𝑑+ 𝑇. The oscillation of a pendulum is an example. The rotation of the earth is another example. (c) The given equation is π‘₯= 𝐴sin(πœ”π‘‘+ 𝛿) Differentiating with respect to time 𝑑, dπ‘₯ d𝑑= π΄πœ”cos(πœ”π‘‘+ 𝛿) Differentiating once more, d2π‘₯ d𝑑2 = βˆ’π΄πœ”2 sin(πœ”π‘‘+ 𝛿) = βˆ’πœ”2π‘₯ d2π‘₯ d𝑑2 + πœ”2π‘₯= 0 The equation satisfies the differential equation. Hence it is a solution of simple harmonic motion. (d) A proper so

Solution of Shamim. Here π‘₯= 𝐡cos(πœ”π‘‘+ 𝛼). dπ‘₯ d𝑑= βˆ’π΅πœ”sin(πœ”π‘‘+ 𝛼) d2π‘₯ d𝑑2 = βˆ’π΅πœ”2 cos(πœ”π‘‘+ 𝛼) = βˆ’πœ”2π‘₯ d2π‘₯ d𝑑2 + πœ”2π‘₯= 0 This solution satisfies the differential equation. Solution of Rimjhim. Here π‘₯= 𝐢tan(πœ”π‘‘+ πœ‘). dπ‘₯ d𝑑= πΆπœ”sec2(πœ”π‘‘+ πœ‘) d2π‘₯ d𝑑2 = 2πΆπœ”2 sec2(πœ”π‘‘+ πœ‘) tan(πœ”π‘‘+ πœ‘) = 2πœ”2π‘₯sec2(πœ”π‘‘+ πœ‘) The right side is not equal to βˆ’πœ”2π‘₯. This solution does not satisfy the differential equation. πœ‹ 2πœ‹ 3πœ‹ 4πœ‹ βˆ’π΅ 𝐡 cosine tangent πœ”π‘‘ π‘₯ The cosine curve stays between βˆ’π΅and +𝐡. The tangent curve becomes infinite again and again. A real oscillating body cannot have an infinite displacement. The solution of Shamim is proper. The solution of Rimjhim is not proper. ahbab.dev 2

Question 2 A particle is revolving around the point 𝑂in a circular path 𝐴𝐡𝐢𝐷of radius 0.02 m along the clockwise direction with an angular velocity 1 rad sβˆ’1 and in 10 s it comes to 𝑃from 𝐴. 𝑋 π‘Œ 𝐴 𝐡 𝐢 𝐷 𝑂 𝑃 𝑁 πœƒ πœƒ πœ” π‘₯ (a) What is called simple harmonic motion? (b) Explain the relation of force constant and mass in case of a spring. (c) Deduce the differential equation of the motion along the 𝑋-axis of the particle of the stimulus. (d) If the initial phase of the particle is 0, then what is the displacement at 𝑃along the 𝑋-axis? Solution (a) In simple harmonic motion, the acceleration is proportional to the displacement. It is always directed towards the equilibrium position. (b) Let a body of mass π‘šoscillate on a spring of force constant π‘˜. πœ”= βˆšοΈ‚ π‘˜ π‘š, 𝑇= 2πœ‹ βˆšοΈ‚π‘š π‘˜ A larger force constant means a stiffer spring. The restoring force is then greater and the time period is smaller. A larger mass has more inertia. The time period is then greater. For a fixed time period, π‘˜is proportional to π‘š. (c) Let the radius of the circle be 𝑅. The particle starts from 𝐴and moves clockwise. In a time 𝑑it turns through an angle πœƒ= πœ”π‘‘from 𝑂𝐴. The foot of the perpendicular from 𝑃on the 𝑋-axis is 𝑁. In th

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