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A simple pendulum is mounted in an elevator. What happens to the period of the pendulum (does it increase, decrease, or remain the same) if the elevator (a) accelerates upward at 5.0 m/s2; (b) moves upward at a steady 5.0 m/s ; (c) accelerates downward at 5.0 m/s2; (d) accelerates downward at 9.8 m/s2Justify your answers.

User Mick Bruno
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1 Answer

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Answer:

(a) Decrease

(b) Will remain same

(c) Increase

(d) Time period will be infinite

Step-by-step explanation:

Time period of simple pendulum is given as
T=2\pi \sqrt{(l)/(g)}

From he expression we can see that time period is inversely proportional to the acceleration due to gravity

(a) Upward acceleration is
5m/sec^2

So
g_(net)=g+5

As the acceleration due to gravity increases so time period will decrease

(b) Moves upward with at a steady 5 m /sec

So
g_(net)=g

So time period will be same

(c) Downward acceleration is
5m/sec^2

So
g_(net)=g-5

As the acceleration due to gravity decreases so time period will increase

(d) Downward acceleration is
9.8m/sec^2

So
g_(net)=g-9.8=0

As the acceleration due to gravity IS 0 so time period will be infinite

User Ethan Hohensee
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