HSC Resources Β· Physics 1
Lecture slides and notes for Ch08 Periodic Motion from the Physics 1 module in HSC Resources by Md Ahbab. 14 pages.

PHYSICS 1ST PAPER CHAPTER 8 Periodic Motion Question 1 d2π₯ dπ‘2 + π2π₯= 0 is a differential equation of a simple harmonic motion. When asked to write its solution, Shamim wrote π₯= π΅cos(ππ‘+ πΌ) and Rimjhim wrote π₯= πΆtan(ππ‘+ π). (a) What is called periodic motion? (b) What do you mean by temporal periodicity? (c) Show that π₯= π΄sin(ππ‘+ πΏ) 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 π. The state of the body at a time π‘returns at the time π‘+ π. The oscillation of a pendulum is an example. The rotation of the earth is another example. (c) The given equation is π₯= π΄sin(ππ‘+ πΏ) Differentiating with respect to time π‘, dπ₯ dπ‘= π΄πcos(ππ‘+ πΏ) Differentiating once more, d2π₯ dπ‘2 = βπ΄π2 sin(ππ‘+ πΏ) = βπ2π₯ d2π₯ dπ‘2 + π2π₯= 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₯= π΅cos(ππ‘+ πΌ). dπ₯ dπ‘= βπ΅πsin(ππ‘+ πΌ) d2π₯ dπ‘2 = βπ΅π2 cos(ππ‘+ πΌ) = βπ2π₯ d2π₯ dπ‘2 + π2π₯= 0 This solution satisfies the differential equation. Solution of Rimjhim. Here π₯= πΆtan(ππ‘+ π). dπ₯ dπ‘= πΆπsec2(ππ‘+ π) d2π₯ dπ‘2 = 2πΆπ2 sec2(ππ‘+ π) tan(ππ‘+ π) = 2π2π₯sec2(ππ‘+ π) The right side is not equal to βπ2π₯. This solution does not satisfy the differential equation. π 2π 3π 4π βπ΅ π΅ cosine tangent ππ‘ π₯ The cosine curve stays between βπ΅and +π΅. 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 πin a circular path π΄π΅πΆπ·of radius 0.02 m along the clockwise direction with an angular velocity 1 rad sβ1 and in 10 s it comes to πfrom π΄. π π π΄ π΅ πΆ π· π π π π π π 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 π-axis of the particle of the stimulus. (d) If the initial phase of the particle is 0, then what is the displacement at πalong the π-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 πoscillate on a spring of force constant π. π= βοΈ π π, π= 2π βοΈπ 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, πis proportional to π. (c) Let the radius of the circle be π . The particle starts from π΄and moves clockwise. In a time π‘it turns through an angle π= ππ‘from ππ΄. The foot of the perpendicular from πon the π-axis is π. In th
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