Physics 1st Ch 08 Periodic Motion

Physics – First Paper Chapter 8: Periodic Motion Curated Questions and Complete Model Solutions (HSC English Version) Chapter 8: Periodic Motion B. Creative Questions 1. Salam fixed a spring of length L to the ceiling of a room and attached a mass of 50 g at its lower end. The length of the spring increased by l = 4.9 cm. He then pulled the mass slightly downward and released it. The spring started oscillating up and down with the mass. [g = 9.8 m/s2] (a) What is simple harmonic motion? (b) Mention two essential characteristics of simple harmonic motion. (c) Show that the vertical oscillation of the spring is simple harmonic motion. (d) If a mass of 200 g is attached to the spring instead of 50 g, what change will occur in its time period? Analyze mathematically. Answer: (a) Simple Harmonic Motion (SHM): If the acceleration of an oscillating body is directly proportional to its displacement from the equilibrium position and is always directed toward that equilibrium position, its motion is called simple harmonic motion. (b) Characteristics of SHM: 1. The motion is periodic and oscillatory. 2. The acceleration is directly proportional to the displacement (a ∝−x) and is always direct

Because the acceleration is directly proportional to the displacement and directed opposite to it (toward the mean position), the vertical oscillation of the spring is a simple harmonic motion. (d) Mathematical analysis of the change in time period: The time period of a mass-spring system is: T = 2π rm k For the initial mass m1 = 50 g = 0.05 kg: T1 = 2π rm1 k For the new mass m2 = 200 g = 0.20 kg: T2 = 2π rm2 k Taking the ratio of T2 to T1: T2 T1 = rm2 m1 = r 200 50 = √ 4 = 2 Thus, T2 = 2T1. Let us calculate the numerical values: k = m1g l = 0.05 × 9.8 0.049 = 10 N/m T1 = 2π r 0.05 10 = 2π √ 0.005 ≈0.444 s T2 = 2 × 0.444 = 0.888 s The time period doubles, increasing by ∆T = T2 −T1 = 0.444 s (a 100% increase). 2. The mass of the bob of the simple pendulum shown in the figure is 50 g and its effective length is L = 1 m. L (a) What is a simple pendulum? (b) What is meant by the effective length of a simple pendulum? (c) Explain whether the pendulum of the stimulus will run fast or slow if it is taken to the Moon. (d) If the mass of the bob is increased to 100 g while the effective length remains unchanged, what change in the time period will take place? Show by mathematical analysis.

Answer: (a) Simple Pendulum: An idealized simple pendulum consists of a heavy point mass sus- pended from a rigid, frictionless support by a weightless, inextensible, and completely flexible string. (b) Effective Length: The effective length (L) of a simple pendulum is the distance from the point of suspension to the center of gravity of the bob. If l is the length of the string and r is the radius of the spherical bob, then: L = l + r (c) Behavior on the Moon: The time period of a simple pendulum is given by: T = 2π s L g On the surface of the Moon, the acceleration due to gravity is approximately one-sixth of that on Earth (gm ≈1 6ge). Since T ∝ 1 √g, as g decreases, the time period T increases. A larger time period means the pendulum takes more time to complete each oscillation; hence, the pendulum will swing more slowly and the clock will run slow. (d) Mathematical analysis of changing bob mass: The formula for the time period of a simple pendulum is: T = 2π s L g Notice that the mass of the bob (m) does not appear anywhere in this formula. When the mass of the bob is increased from m1 = 50 g to m2 = 100 g, the effective length L and the local gravitational acceleration g remai