The period of oscillation of a simple pendulum of constant length at surface of the earth is T. Its time period inside a mine will be :

Question

The period of oscillation of a simple pendulum of constant length at surface of the earth is T. Its time period inside a mine will be : (A) cannot be compared. (B) equal to T (C) less than T (D) more than T

Tardigrade Admin · Accepted Answer

Time period of a simple pendulum at the surface of earth, T=2π √(l/g) T propto (1/√g) In a mine (below the surface of earth) value of 'g' will decrease, so time period 'T' will increase.

Q. The period of oscillation of a simple pendulum of constant length at surface of the earth is T. Its time period inside a mine will be :

cannot be compared.

equal to T

less than T

more than T

Time period of a simple pendulum at the surface of earth, $T = 2 π \frac{l}{g}$ $T \propto \frac{1}{g}$ In a mine (below the surface of earth) value of 'g' will decrease, so time period 'T' will increase.

Q. The period of oscillation of a simple pendulum of constant length at surface of the earth is T. Its time period inside a mine will be :

cannot be compared.

equal to T

less than T

more than T

Time period of a simple pendulum at the surface of earth, T=2πgl​​ T∝g​1​ In a mine (below the surface of earth) value of 'g' will decrease, so time period 'T' will increase.

Time period of a simple pendulum at the surface of earth, $T = 2 π \frac{l}{g}$ $T \propto \frac{1}{g}$ In a mine (below the surface of earth) value of 'g' will decrease, so time period 'T' will increase.