George W. answered • 01/04/21
Physical Science Blogger; Stock Options Trader; AP Physics Tutor
This question regards the period ( T ) of a pendulum. The period ( T ) is defined as the time ( t ) that passes as a pendulum completes 1 revolution. Objects revolving back and forth are quantified as having an angular speed.
Imagine an object rotating along the circumference ( C ) of a circle. Imagine that the object is positioned above the x-axis of a graph. If the rotating object has a " shadow " that falls upon the x-axis, this shadow would move back and forth with the same frequency ( f ) and period ( T ) of the object moving in angular motion on the circle.
Angular motion is defined as follows: For any unit circle, ( l / r ) = theta, where l = a length along the circumference of a circle, r = the radius of the circle, and theta = the angle that corresponds to ( l / r ). Thus, l = ( r )( theta ). If we divide both sides of the equation by seconds ( s ), we have the linear component of velocity ( v ) in meters per second that an object travels around the circle, and we have the corresponding angular velocity as well. Therefore, v = ( r )( w ), where w = ( theta / s ), where theta is in radians.
A pendulum accelerates centripetally around a central point of reference. Therefore, the expression of v = ( r )( w ) must be algebraically converted to centripetal acceleration a = ( v^2 / r ). Thus, v^2 = ( r^2 )( w^2 ). Dividing both sides of the equation by r gives us a = ( v^2 / r ) = ( r )( w^2 ).
The value of r in the above example regards the length ( L ) of the pendulum; it is the " radius " of our system of interest. Furthermore, acceleration due to gravity ( g ) drives the pendular system, so a = g = ( 9.8 m / s^2 ). As a consequence, g = ( L )( w^2 ). Recall that w = angular speed in radians per second ( rad / s ). The value of radians per second = frequency ( f ). The inverse of frequency ( f ) is the period ( T ) in seconds per radian. In other words, T gives us the time ( t ) needed to complete a revolution, and this is what is asked for in the question. Since f = ( 1 / T ), and w ( radians / sec ) = f, we can first solve for frequency ( f ). Since ( 1 / f ) = T, the inverse of f will give us the answer we seek.
Solving for w, we have g = ( L )( w^2 ). Since ( g / L ) = w^2, w = the square root ( sqrt ) of ( g / L ). It can also be shown that w = ( 2 * pi * f ), because one complete revolution is made when 2*pi radians have moved through a reference point is one second. Therefore, w = ( 2 * pi * f ), and the sqrt of ( g / L ) = ( 2 * pi * f ). Solving for frequency gives us f = ( sqrt g / L ) )( 1 / 2 pi ). Taking the invese of f gives us T, so ( 1 / f ) = T = ( sqrt L / g )( 2 pi ).
T = ( sqrt L / g )( 2 pi ) = ( sqrt ( 2.45 m / 9.8 m / s^2 ) )( 2 pi ) = ( 0.5 s )( 2 pi ) = 3.14 s.
Best of luck in your studies!
