Pendulums are in common usage. Grandfather clocks use a pendulum to keep time and a pendulum can be used to measure the acceleration due to gravity. For small displacements, a pendulum is a simple harmonic oscillator.
A simple pendulum is defined to have a point mass, also known as the pendulum bob , which is suspended from a string of length L with negligible mass (Figure 15.20). Here, the only forces acting on the bob are the force of gravity (i.e., the weight of the bob) and tension from the string. The mass of the string is assumed to be negligible as compared to the mass of the bob.
Figure 15.20 A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is s, the length of the arc. Also shown are the forces on the bob, which result in a net force of − m g sin θ − m g sin θ toward the equilibrium position—that is, a restoring force.
Consider the torque on the pendulum. The force providing the restoring torque is the component of the weight of the pendulum bob that acts along the arc length. The torque is the length of the string L times the component of the net force that is perpendicular to the radius of the arc. The minus sign indicates the torque acts in the opposite direction of the angular displacement:
τ = − L ( m g sin θ ) ; I α = − L ( m g sin θ ) ; I d 2 θ d t 2 = − L ( m g sin θ ) ; m L 2 d 2 θ d t 2 = − L ( m g sin θ ) ; d 2 θ d t 2 = − g L sin θ . τ = − L ( m g sin θ ) ; I α = − L ( m g sin θ ) ; I d 2 θ d t 2 = − L ( m g sin θ ) ; m L 2 d 2 θ d t 2 = − L ( m g sin θ ) ; d 2 θ d t 2 = − g L sin θ .
The solution to this differential equation involves advanced calculus, and is beyond the scope of this text. But note that for small angles (less than 15 degrees or about 0.26 radians), sin θ sin θ and θ θ differ by less than 1%, if θ is measured in radians. We can then use the small angle approximation sin θ ≈ θ . sin θ ≈ θ . The angle θ θ describes the position of the pendulum. Using the small angle approximation gives an approximate solution for small angles,
d 2 θ d t 2 = − g L θ . d 2 θ d t 2 = − g L θ .Because this equation has the same form as the equation for SHM, the solution is easy to find. The angular frequency is
ω = g L ω = g Land the period is
T = 2 π L g . T = 2 π L g .The period of a simple pendulum depends on its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass and the maximum displacement. As with simple harmonic oscillators, the period T for a pendulum is nearly independent of amplitude, especially if θ θ is less than about 15 ° . 15 ° . Even simple pendulum clocks can be finely adjusted and remain accurate.
Note the dependence of T on g. If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity, as in the following example.
What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?
We are asked to find g given the period T and the length L of a pendulum. We can solve T = 2 π L g T = 2 π L g for g, assuming only that the angle of deflection is less than 15 ° 15 ° .
This method for determining g can be very accurate, which is why length and period are given to five digits in this example. For the precision of the approximation sin θ ≈ θ sin θ ≈ θ to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about 0.5 ° 0.5 ° .
An engineer builds two simple pendulums. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg. Pendulum 2 has a bob with a mass of 100 kg. Describe how the motion of the pendulums will differ if the bobs are both displaced by 12 ° 12 ° .
Any object can oscillate like a pendulum. Consider a coffee mug hanging on a hook in the pantry. If the mug gets knocked, it oscillates back and forth like a pendulum until the oscillations die out. We have described a simple pendulum as a point mass and a string. A physical pendulum is any object whose oscillations are similar to those of the simple pendulum, but cannot be modeled as a point mass on a string, and the mass distribution must be included into the equation of motion.
As for the simple pendulum, the restoring force of the physical pendulum is the force of gravity. With the simple pendulum, the force of gravity acts on the center of the pendulum bob. In the case of the physical pendulum, the force of gravity acts on the center of mass (CM) of an object. The object oscillates about a point O. Consider an object of a generic shape as shown in Figure 15.21.
an irregularly shaped object. The center of mass, C M, is a distance L from the pivot point, O. The center of mass traces a circular arc, centered at O. The line from O to L makes an angle theta counterclockwise from the vertical. Three forces are depicted by red arrows at the center of mass. The force m g points down. Its components are minus m g sine theta which points tangent to the arc traced by the center of mass, and m g cosine theta which points radially outward." width="381" height="382" />
Figure 15.21 A physical pendulum is any object that oscillates as a pendulum, but cannot be modeled as a point mass on a string. The force of gravity acts on the center of mass (CM) and provides the restoring force that causes the object to oscillate. The minus sign on the component of the weight that provides the restoring force is present because the force acts in the opposite direction of the increasing angle θ θ .
When a physical pendulum is hanging from a point but is free to rotate, it rotates because of the torque applied at the CM, produced by the component of the object’s weight that acts tangent to the motion of the CM. Taking the counterclockwise direction to be positive, the component of the gravitational force that acts tangent to the motion is − m g sin θ − m g sin θ . The minus sign is the result of the restoring force acting in the opposite direction of the increasing angle. Recall that the torque is equal to τ → = r → × F → τ → = r → × F → . The magnitude of the torque is equal to the length of the radius arm times the tangential component of the force applied, | τ | = r F sin θ | τ | = r F sin θ . Here, the length L of the radius arm is the distance between the point of rotation and the CM. To analyze the motion, start with the net torque. Like the simple pendulum, consider only small angles so that sin θ ≈ θ sin θ ≈ θ . Recall from Fixed-Axis Rotation on rotation that the net torque is equal to the moment of inertia I = ∫ r 2 d m I = ∫ r 2 d m times the angular acceleration α , α , where α = d 2 θ d t 2 α = d 2 θ d t 2 :
I α = τ net = L ( − m g ) sin θ . I α = τ net = L ( − m g ) sin θ .Using the small angle approximation and rearranging:
I α = − L ( m g ) θ ; I d 2 θ d t 2 = − L ( m g ) θ ; d 2 θ d t 2 = − ( m g L I ) θ . I α = − L ( m g ) θ ; I d 2 θ d t 2 = − L ( m g ) θ ; d 2 θ d t 2 = − ( m g L I ) θ .
Once again, the equation says that the second time derivative of the position (in this case, the angle) equals minus a constant ( − m g L I ) ( − m g L I ) times the position. The solution is
θ ( t ) = Θ cos ( ω t + ϕ ) , θ ( t ) = Θ cos ( ω t + ϕ ) ,where Θ Θ is the maximum angular displacement. The angular frequency is
