A simple pendulum with a period T = 2.50 s is attached to the ceiling of an elevator which is initially at rest.

a. Calculate the period of oscillations when the elevator begins to accelerate upwards with acceleration a = 1.50 m/s^2.
b. Calculate the period of oscillations when the elevator accelerates downwards with acceleration a = 2.50 m/s^2 .

Explanation:

For a simple harmonic motion, the period, T = 2π ∙ √( L / g )

When the elevator is at rest:

2.5 = 2π * √( L/9.81)

L = 1.5515 m.

B.

When the elevator is accelerating upwards at 1.5 m/s²

T = 2π ∙ √( 1.5515/(9.81 + 1.5))

= 2.33 s.

B.

When the elevator is accelerating downwards at 2.5 m/s²

T = 2π * √(1.5515/(9.81 - 2.5))

T = 2.89 s.

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west