A simple pendulum is 5.00 m long. (a) What is the period of simple harmonic moti

Question

A simple pendulum is 5.00 m long. (a) What is the period of simple harmonic motion for this pendulum if it is located in an elevator accelerating upward at 8.00 m/s2?
s

(b) What is its period if the elevator is accelerating downward at 8.00 m/s2?
s

(c) What is the period of simple harmonic motion for the pendulum if it is placed in a truck that is accelerating horizontally at 8.00 m/s2?
s (a) What is the period of simple harmonic motion for this pendulum if it is located in an elevator accelerating upward at 8.00 m/s2?
s

(b) What is its period if the elevator is accelerating downward at 8.00 m/s2?
s

(c) What is the period of simple harmonic motion for the pendulum if it is placed in a truck that is accelerating horizontally at 8.00 m/s2?
s

Explanation / Answer

a) What is the period of simple harmonic motion for this pendulum if it is hanging in an elevator that is accelerating upward at 8.00 m/s2?

T = 2x3.14x sqrt[5/(9.8+8)] = 3.33 s

(b) What is its period if the elevator is accelerating downward at 8.00 m/s2?
T = 2x3.14x sqrt[5/(9.8-8)] = 10.47 s

(c) What is the period of simple harmonic motion for this pendulum if it is placed in a truck that is accelerating horizontally at 8.00 m/s2?

I guess that the effective length would reduce to 5 cos theta

T cos theta = mg and T sin theta = ma or theta = tan ^-1 (a/g)

theta = tan^-1(8/9.8) = 39.23

Effective length = 5 cos theta = 3.87 m

Hence T = 2x3.14xsqrt[3.87/9.8] = 3.95 s

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