HSC Resources Β· Physics Homework Notes
Lecture slides and notes for Ch08 Periodic Motion from the Physics Homework Notes 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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