What is simple harmonic motion derive differential equation of SHM?

Simple Harmonic Motion can be used to describe the motion of a mass at the end of a linear spring without a damping force or any other outside forces acting on the mass. It's best thought of as the motion of a vibrating spring.

Laws of Motion[edit | edit source]

There are generally two laws that help describe the motion of a mass at the end of the spring.

1. Hooke's Law
2. Newton's Second Law

Hooke’s Law[edit | edit source]

To demonstrate Hooke’s Law, we will use a (massless) spring hung from a ceiling. The ceiling is rigid, and offers no effect on the spring’s motion.

If we leave the spring alone, and don’t attach a mass, the spring will remain in an un-stretched state with a total length l {\displaystyle l}

m 1 > 0 {\displaystyle m_{1}>0}

s {\displaystyle s}

length total = s + l {\displaystyle {\mbox{length}}_{\mbox{total}}=s+l}

Eventually, the mass will come to rest at this new total length at a position known as the Equilibrium Position. This position is what many of the calculation reference as x = 0 {\displaystyle x=0}

Here is where Hooke's law comes into play. As the spring stretches, when the mass is attached, a force is exerted on the mass in the direction of the original un-stretched position. This force is expressed using the equation: F → = k → s {\displaystyle {\vec {F}}={\vec {k}}s}

F → {\displaystyle {\vec {F}}}

k → {\displaystyle {\vec {k}}}

Suppose that any movement by the mass in a downward direction is considered positive and upward is negative. If you take the weight ( W → {\displaystyle {\vec {W}}}

W → = F → {\displaystyle {\vec {W}}={\vec {F}}}

W → = m g → {\displaystyle {\vec {W}}=m{\vec {g}}}

m g → = k → s {\displaystyle m{\vec {g}}={\vec {k}}s}

m g → − k → s = 0 {\displaystyle m{\vec {g}}-{\vec {k}}s=0}

Why is k → s {\displaystyle {\vec {k}}s}

Examples[edit | edit source]

Example 1[edit | edit source]

A mass of 1 / 4  slug {\displaystyle 1/4{\mbox{ slug}}}

6  inches {\displaystyle 6{\mbox{ inches}}}

Solution
Using the combined equation m g − k s = 0 {\displaystyle mg-ks=0}

m = 1 / 4  slug {\displaystyle m=1/4{\mbox{ slug}}}

s = 6  inches = 1 / 2  ft {\displaystyle s=6{\mbox{ inches}}=1/2{\mbox{ ft}}}

(we omit the vector since we're only looking for a magnitude)

By definition, we know g = 32 f t / s 2 {\displaystyle g=32ft/s^{2}}

Plugging everything in, we get ( 1 / 4 s l u g ) ( 32 f t / s 2 ) − k ( 1 / 2 f t ) = 0 {\displaystyle (1/4slug)(32ft/s^{2})-k(1/2ft)=0}

Solving for k {\displaystyle k}

8  lb = k ( 1 / 2  ft ) {\displaystyle 8{\mbox{ lb}}=k(1/2{\mbox{ ft}})}

16  lb / ft = k {\displaystyle 16{\mbox{ lb}}/{\mbox{ft}}=k}

Example 2[edit | edit source]

An object of unknown mass stretches a spring 10 cm from the ceiling. The spring's original length was 7 cm. Calculate the weight of the object if the spring constant is 5 N/m.

Solution
First, we need the distance the spring is stretched after the mass is attached. This can be found using length total = s + l {\displaystyle {\mbox{length}}_{\mbox{total}}=s+l} , where length total = 10  cm = .1  m {\displaystyle {\mbox{length}}_{\mbox{total}}=10{\mbox{ cm}}=.1{\mbox{ m}}}

l = 7  cm = .07  m {\displaystyle l=7{\mbox{ cm}}=.07{\mbox{ m}}}

s = .03  m {\displaystyle s=.03{\mbox{ m}}}

By definition we know W = m g {\displaystyle W=mg}

W {\displaystyle W}

W − ( 5  N / m ) ( .03  m ) = 0 {\displaystyle W-(5{\mbox{ N}}/{\mbox{m}})(.03{\mbox{ m}})=0}

W = .15  N = 150  g {\displaystyle W=.15{\mbox{ N}}=150{\mbox{ g}}}

Newton’s Second Law of Motion[edit | edit source]

Now, imagine that we have another mass hanging from a spring, which is attached to the ceiling. This mass is then pulled down a distance x {\displaystyle x}

If the mass continues this motion, without any outside influence, it's known as Free Motion. During this motion, there is acceleration acting on the mass to keep it in motion. This acceleration can be readily found in Newton's Second Law of Motion using F → = m a → {\displaystyle {\vec {F}}=m{\vec {a}}}

m {\displaystyle m}

a → {\displaystyle {\vec {a}}}

From Calculus, we know that a = d 2 x d t 2 {\displaystyle a={\frac {d^{2}x}{dt^{2}}}}

a = x ¨ {\displaystyle a={\ddot {x}}}

m x ¨ − k ( s + x ) = 0 {\displaystyle m{\ddot {x}}-k(s+x)=0}

m x ¨ + k ( s + x ) = m g {\displaystyle m{\ddot {x}}+k(s+x)=mg}

m x ¨ {\displaystyle m{\ddot {x}}}

m x ¨ = − k ( s + x ) + m g = − k s − k x + m g {\displaystyle m{\ddot {x}}=-k(s+x)+mg=-ks-kx+mg}

m x ¨ = − k x {\displaystyle m{\ddot {x}}=-kx}

The Differential Equation of Free Motion or SHM[edit | edit source]

Finally, if we set the equation above equal to zero, we end up with the following:

m x ¨ + k x = 0 {\displaystyle m{\ddot {x}}+kx=0}

Since our leading coeffiecient should be equal to 1, we divide by the mass to get:

x ¨ + k m x = 0 {\displaystyle {\ddot {x}}+{\frac {k}{m}}x=0}

If we set ω 2 = k m {\displaystyle \omega ^{2}={\frac {k}{m}}}

x ¨ + ω 2 x = 0 {\displaystyle {\ddot {x}}+\omega ^{2}x=0}

The above equation is known to describe Simple Harmonic Motion or Free Motion.

Initial Conditions[edit | edit source]

With the free motion equation, there are generally two bits of information one must have to appropriately describe the mass's motion.

1. The starting position of the mass. x 2 {\displaystyle x_{2}}
   
2. The starting direction and magnitude of motion. v {\displaystyle v}
   

Generally, one isn't present without the other. For simplicity, we will consider all displacement below the equilibrium point as x > 0 {\displaystyle x>0}

x < 0 {\displaystyle x<0}

For upward motion v < 0 {\displaystyle v<0}

v > 0 {\displaystyle v>0}

Solution[edit | edit source]

Multiplying this equation by x ˙ {\displaystyle {\dot {x}}}

m x ¨ x ˙ + k x x ˙ = 0 {\displaystyle m{\ddot {x}}{\dot {x}}+kx{\dot {x}}=0}

The first and the second addends are exact derivatives, so this equation may be integrated to obtain the following relation:

m x ˙ 2 2 + k x 2 2 = E {\displaystyle m{\frac {{\dot {x}}^{2}}{2}}+k{\frac {x^{2}}{2}}=E}

The first addend of this relation is known as the kinetic energy of the mass and the second — as the potential energy of the spring. The above integral represents the energy conservation law. This is also a first order separable differential equation. It may be rewritten as

d x 2 E m − k m x 2 = ± d t {\displaystyle {\frac {dx}{\sqrt {{\frac {2E}{m}}-{\frac {k}{m}}x^{2}}}}=\pm dt}

The integration of this relation gives

arccos ⁡ x k 2 E = ± k m t + φ {\displaystyle \arccos x{\sqrt {\frac {k}{2E}}}=\pm {\sqrt {\frac {k}{m}}}t+\varphi }

Or, finally rearranging the result, substituting ω = k / m {\displaystyle \omega ={\sqrt {k/m}}}

x = 2 E k cos ⁡ ( ω t + φ ) {\displaystyle x={\sqrt {\frac {2E}{k}}}\cos(\omega t+\varphi )}