Question

How does the time period of a simple pendulum change when it is made to oscillate in a trough of water?​

Answer 1

Answer:$T=2\pi \sqrt{l/g(1-1/η)}$

Explanation:

when the pendulum is in water

T=mg -U

T is the tension in the string

mg is the weight of the bob

U is the upthrust experienced by the bob

By applying F=ma we can deduce the equation

mg - U =ma

where a is the free fall acceleration of the pendulum bob

Vρ₀ηg - Vρ₀g = Vρ₀ηa

a = g(1-1/η) ------------1

ρ₀ is  the density of water

η is the relative density of the bob

if the period of the pendulum bob isT

    $T=2\pi \sqrt{l/g}$

where g is the gravitational acceleration which is the freefall acceleration in the experimental procedures

but in this condition a is the freefall acceleration as derived in 1 above

therefore

$T=2\pi \sqrt{l/g(1-1/η)}$