A simple pendulum is 5.00 m long. What is the period of small oscillations for this pendulum if it is located jn an elevator (a) at rest? (b) Accelerating upward at 5.00 m/s²? (c) Accelerating downward at 5.00 m/s²? (d) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 5.00 m/s²?
Question
A simple pendulum is 5.00 m long. What is the period of small oscillations for this pendulum if it is located jn an elevator (a) at rest? (b) Accelerating upward at 5.00 m/s²? (c) Accelerating downward at 5.00 m/s²? (d) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 5.00 m/s²?
Solution
(a) The period of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. In this case, L = 5.00 m and g = 9.81 m/s². Substituting these values into the formula gives T = 2π√(5.00/9.81) = 2.25 s.
(b) If the elevator is accelerating upward, the effective gravity is g' = g + a, where a is the acceleration of the elevator. In this case, a = 5.00 m/s², so g' = 9.81 + 5.00 = 14.81 m/s². Substituting this into the formula gives T = 2π√(5.00/14.81) = 1.83 s.
(c) If the elevator is accelerating downward, the effective gravity is g' = g - a. In this case, g' = 9.81 - 5.00 = 4.81 m/s². Substituting this into the formula gives T = 2π√(5.00/4.81) = 3.19 s.
(d) If the truck is accelerating horizontally, this does not affect the period of the pendulum. The pendulum is only affected by forces acting perpendicular to its motion, so the horizontal acceleration of the truck is irrelevant. The period of the pendulum remains as in case (a), T = 2.25 s.
Similar Questions
hat is the period of a simple pendulum that is 1.13-m long in each situation? In an elevator accelerating at 2.41 m/s22.41 m/s2 upward
a) The period of a simple pendulum is T = 2𝜋L/g where L is the length of the pendulum and g is the acceleration due to gravity at the pendulum's location. Thus, if a pendulum has a period of T = 1.9 s on Earth where gEarth = 9.8 m/s2, its length isLEarth = gEarthT24𝜋2 = m/s2 s 2 4𝜋2 = m = cm
As shown in the above figure, a cart is accelerating in the horizontal direction with a constant acceleration of 3–√g3𝑔 and a 50 cm50 cm long simple pendulum is suspended from its roof. Now, find the time period of small oscillations of this pendulum in second about its equilibrium position taking g=π2 m s−2
A man enters a tall tower, needing to know its height. He notes that a long pendulum extends from the ceiling almost to the floor and that its period is 12.0 s.(a) How tall is the tower? m(b) If this pendulum is taken to the Moon, where the free-fall acceleration is 1.67 m/s2, what is the period there? s
A simple pendulum has a period of 3.65 s on the surface of the Earth. If this pendulum were placed on the surface of the Moon, where the gravitational acceleration is 1.62 m/s2m/s2 , what would its period be?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.