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3 years ago

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A simple pendulum is 3.00 m long. (a) What is the period of small oscillations for this pendulum if it is located in an elevator

accelerating upward at 3.00 m/s2? s (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2? s (c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2? s

1 answer:

aliina [53]3 years ago

5 0

Answer:

a,)3.042s

b)4.173s

c)3.281s

Explanation:

For a some pendulum the period in seconds T can be calculated using below formula

T=2π√(L/G)

Where L = length of pendulum in meters

G = gravitational acceleration = 9.8 m/s²

Then we are told to calculate

(a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2?

Since oscillations for this pendulum is located in the elevator that is accelerating upward at 3.00 then

use G = 9.8 + 3.0 = 12.8 m/s²

Period T=2π√(L/G)

T= 2π√(3/12.8)

T=3.042s

b) (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2?

G = 9.8 – 3.0 = 6.8 m/s²

T= 2π√(3/6.8)

T=4.173s

C)(c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2?

Net acceleration is

g'= √(g² + a²)

=√(9² + 3²)

Then period is

T=2π√(3/11)

T=3.281s

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