You pull a simple pendulum of length 0.225m to the side through an angle of 3.50∘ and release it. Part A How much time does it take the pendulum bob to reach its highest speed? Take free fall acceleration to be 9.80 m/s^2 . t =??? Part B How much time does it take if the pendulum is released at an angle of 1.75∘ instead of 3.50∘? t =???

Respuesta :

The period of a pendulum to complete one full oscillation is: T=2pi[sqroot(L/g)]. It reaches its highest speed at the bottom of the pendulum when all potential energy has been converted to kinetic. When it’s at the bottom it has complete one fourth of its oscillation. So the time it takes the pendulum to reach its highest speed is one fourth of T=2pi[sqroot(L/g)]. Now simply plug in the values. It’ll be the same for angle 1.75 since the period does not depend on the displacement of the angle only the length and strength of gravity

The period of a pendulum is the time to complete one oscillation. The period for both the angles will be same which is 0.983 sec.

The period of a pendulum to complete one oscillation is:

[tex]\bold {T=2\pi \sqrt \dfrac Lg}[/tex]

Where,

T - period

L - length = 0.225 m

g - gravitational acceleration = [tex]\bold {9.8 m/s^2}[/tex]

Put the value in the formula,

[tex]\bold {T=2\pi \sqrt \dfrac {0.225\ m}{9.8\ m/s^2 }}\\\\\\\\\\\bold {\ = 0.983\ s}[/tex]

Since, the period of pendulum does not depend up on the angle of displacement.

Therefore, the period for both the angles will be same which is 0.983 sec.

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