The period of a pendulum, given by the equation T=2π √l/ g , provides a real-value period for which of the following domains? a. L>0,g>0 b. L=0,g≠0 c. L≠0,g=0 d L<0,g&l…

Question

The period of a pendulum, given by the equation

T=2π √l/ g , provides a real-value period for which of the following domains?
a. L>0,g>0
b. L=0,g≠0
c. L≠0,g=0
d L<0,g<0

Answer 1

The period of a pendulum equation, T=2π √l/ g, provides a real-value period only for the domain L>0, g>0. This occurs when the pendulum is in simple harmonic motion, which is the case when the angle of deflection is less than 15 degrees.

Step-by-step explanation:

We are asked to determine the real-value period of a pendulum for specific domain conditions. The domain conditions ensure the equation T=2π √l/ g returns a real value. The pendulum equation only provides a real-value period when both the length L and the acceleration due to gravity g are greater than zero. Therefore, the correct domain for which the equation provides a real-value period is L>0, g>0.

When the angle of deflection is less than 15 degrees, a pendulum exhibits simple harmonic motion (SHM), and the equation can be used. In such scenarios, the period T is independent from other factors like mass or amplitude, but it depends on L (length of the pendulum) and g (acceleration due to gravity).

Given the period T and the length L of a pendulum, we can find g by manipulating the formula as follows: g = (4π^2 · L) / T^2, assuming that the simple harmonic motion conditions are met.