Simple harmonic motion:A = amplitudeT = periodω = angular frequencyf = frequencyf = 1/Tω = 2π/T = ω = 2πf

An object whose position as a function of time varies as x(t) = Acos(ωt) exhibits simple harmonic motion.The velocity of the object as a function of time is given by v(t) = -ωAsin(ωt), and the acceleration is given by a(t) = -ω2Acos(ωt) = -ω2x.

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F = -kx = mω2x.ω2 = k/m,ω = √(k/m), T = 2π√(m/k), f = (1/2π)√(k/m).

Simple harmonic motion is accelerated motion. If an object exhibits simple harmonic motion if it is acted on by a restoring force F = -kx, with k = mω2. A mass oscillating on a spring is an example of an object moving with simple harmonic motion.

### The pendulum: Most system which have an equilibrium position execute simple harmonic motion about this position when they are displaced from equilibrium, as long as the displacements are small. The restoring forces approximately obey Hooke"s law. However, for larger displacements the systems becomeanharmonic oscillators, i.e.the restoring forces are no longer proportional to lớn the displacements. The period then depends on the amplitude. A familiar example of such a system is the simple pendulum. An ideal simple pendulum consists of a point mass m suspended from a tư vấn by a massless string of length L. (A good approximation is a small mass, for example a sphere with a diameter much smaller than L, suspended from a light string.) The equilibrium position of the mass is a distance L below the support. The net force acting on m is F=-mgsinθ. It is a restoring force. The displacement from the equilibrium position is s = Lθ. When the displacement from equilibrium is small, then sinθ ~ θ in radians. (See graph.)

Then F = -mgθ = -(mg/L)s. The force obeys Hooke"s law, F = -ks, with k = mg/L.

For small displacements s the pendulum executes simple harmonic motion s(t) = smaxcos(ωt + φ), with ω2 = g/L.

For small oscillations the period of a simple pendulum therefore is given by T = 2π/ω = 2π√(L/g).It is independent of the mass m of the bob. It depends only on the strength of the gravitational acceleration g & the length of the string L. By measuring the length and the period of a simple pendulum we can determine g.

Problem:

Simple pendulum:

F = -mgsinθ,s = Lθ.

For small displacements:

F = -mgθ = -(mg/L)s,k = (mg/L),ω2 = k/m = g/L,T = 2π√(L/g), f = 1/T.

The angular displacement of a pendulum is represented by the equation θ = 0.32*cos(ωt) where θ is in radians và ω = 4.43 rad/s. Determine the period & length of the pendulum.

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Solution:

Reasoning:θ(t) = θmaxcos(ωt + φ) for small oscillations. Details of the calculation:Here ω = 4.43/s, ω2 = g/L = 19.62/s2, L = 0.5 m.T = 2π/ω = 1.42 sProblem:

A simple pendulum on a cuckoo clock is 5.00 cm long. What is its frequency?

Solution:

Reasoning:For a simple pendulum ω2 = g/L.Details of the calculation:f = 1/T = ω/(2π) = (g/L)1/2/(2π) = 2.23/s.Problem:

A simple pendulum with a period of 2.00000 s in one location where g = 9.80 m/s2 is moved khổng lồ a new location where the period is now 1.99796 s. What is the acceleration due lớn gravity at its new location?

Solution:

Reasoning:For a simple pendulum the period T = 2π)/ω is proportional khổng lồ 1/g1/2.Details of the calculation:Told/Tnew = (gnew/gold)1/2. Gnew = gold*(Told/Tnew) = 9.8*(2/1.99796)2 = 9.82 m/s2.Problem:

Pendulum clocks are made lớn run at the correct rate by adjusting the pendulum"s length. Suppose you move from one city to another where the acceleration due to lớn gravity is slightly greater, taking your pendulum clock with you, will you have to lớn lengthen or shorten the pendulum to keep the correct time, other factors remaining constant?

Solution:

Reasoning:For a simple pendulum the period T is proportional lớn (L/g)1/2. If g increases slightly, L has to increase slightly to keep the period the same. You will have khổng lồ lengthen the pendulum.