Question

a student standing in an elevator at rest notices that her pendulum has a frequency of oscillation of 0.5 Hz. a. what is the length of the pendulum's string? b. the elevator starts to accelerate upwards. what effect will this have on the oscillation? explain.​

Answer 1

Step-by-step explanation:

Given that,

The frequency of oscillation, f = 0.5 Hz

(a) Let the length of the pendulum string is l. So,

`T=2\pi \sqrt{(l)/(g)}`

T is time period, T = 1/f

So,

`l=(((1)/(f))^2g)/(4\pi ^2)\\\\l=(((1)/(0.5))^2* 9.8)/(4* 3.14 ^2)\\\\l=0.993\ m`

(b) When the elevator starts to accelerate upwards, new acceleration is a+g. If accelerating upward, the value of g has increased, because of inertia. So it will feel heavier and the period will decrease.