OSCILLATIONS AND WAVES: The Period of a Pendulum

Q: A pendulum has a length of 2.45 m. How many seconds will it take the pendulum to swing through one complete cycle? Does the mass of the pendulum influence the period? Why or why not?

A: This question regards the period ( T ) of a pendulum. The period ( T ) is defined as the time ( t ) that passes as a pendulum moves through a complete revolution. The back-and-forth angular speed of any oscillating object can be quantified with the same mathematics. In the image below, an object rotates along the circumference ( C ) of a circle. If the rotating object had a ” shadow ” on 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 along the circle.

For any unit circle, ( l / r ) = θ. The length ( l ) in question is the distance traced along the circumference of a circle by a radius ( r ) that has moved some counterclockwise angular distance ( θ ) from the x-axis. Thus, l = ( r )( θ ). By dividing both sides of the equation by seconds ( s ), we derive v = ( r )( ω ). The angular speed ( ω ) is measured in radians per second = ( θ / s ). A radian is the length along the circumference of a circle that is equal to the circle’s radius ( r ). The circumference of a circle contains 6.28 radians ( 2𝜋 ).

A pendulum accelerates centripetally around a central point of reference. Therefore, the expression of v = ( r )( ω ) must be algebraically converted to centripetal acceleration a_c = ( v² / r ). Thus, v² = ( r² )( ω² ). Dividing both sides of the equation by r gives us a_c = ( v² / r ) = ( r )( ω² ). When dealing with pendulums, the radius ( r ) is the length of the pendulum. Furthermore, acceleration due to gravity ( g ) drives the pendular system, so a_c = g = ( 9.8 m / s² ). As a consequence, g = ( L )( ω² ). 

Since ( g / L ) = ω², the angular speed ( ω ) = √ ( g / L ). Since angular speed is measured in radians per second, ω = 2𝜋f. The SI unit of frequency ( f ) is Hertz ( Hz ), and it’s units are cycles per second. Therefore, the √ ( g / L ) = 2𝜋f. Solving for frequency gives us f = √ ( g / L )( 1 / 2𝜋 ). Finally,( 1 / f ) = T = √ ( L / g )( 2𝜋).

T = √ ( L / g )( 2𝜋 ) = ( √ ( 2.45 m / 9.8 m / s² ) )( 2𝜋 ) = ( 0.5 s )( 2𝜋 ) = 3.14 s.