The Student Room Group

Derive time period for spring in SHM

Derive T=2Pi(m/k)^1/2

I've got as far as T=2Pi/w => 1/w=(m/k)^1/2 => w^2=k/m
I don't even know what k/m is.
I've tried integrating trig graphs, rearranging spring constant equations, all to no avail.
If anyone could give me a pointer that would be much appreciated.
Reply 1
Original post by TheSK00T3R
Derive T=2Pi(m/k)^1/2

I've got as far as T=2Pi/w => 1/w=(m/k)^1/2 => w^2=k/m
I don't even know what k/m is.
I've tried integrating trig graphs, rearranging spring constant equations, all to no avail.
If anyone could give me a pointer that would be much appreciated.


The defining feature of SHM is that axa \propto -x where xx is the displacement from the origin. We usually write the associated equation in the form a=ω2xa=-\omega^2 x, because then after solving, we find that the period is given by T=2πωT=\frac{2\pi}{\omega}.

For a mass on a spring, the restoing force is given by F=ma=kxF = ma = -kx where kk is the spring constant.

Can you finish it from here? (Hint: rearrange and equate coefficients)
Reply 2
Ah right:
ma=kx[br]a=kxm[br]a=ω2x[br]kxm=ω2x=>kxmx=ω2=>w=kmma=-kx[br]a=\dfrac{-kx}{m}[br]a=-\omega^2x[br]\dfrac{-kx}{m}=-\omega^{2}x => \dfrac{-kx}{mx}=-\omega^{2}=>w=\sqrt{\dfrac{k}{m}}
sub ω\omega back in:
[br]T=2πkm=2πmk[br][br]T=\dfrac{2\pi}{\sqrt{\dfrac{k}{m}}}={2\pi}\sqrt{\dfrac{m}{k}} [br]
Reply 3
Ah right:
ma=kx[br]a=kxm[br]a=ω2x[br]kxm=ω2x=>kxmx=ω2=>w=kmma=-kx[br]a=\dfrac{-kx}{m}[br]a=-\omega^2x[br]\dfrac{-kx}{m}=-\omega^{2}x => \dfrac{-kx}{mx}=-\omega^{2}=>w=\sqrt{\dfrac{k}{m}}
sub ω\omega back in:
[br]T=2πkm=2πmk[br][br]T=\dfrac{2\pi}{\sqrt{\dfrac{k}{m}}}={2\pi}\sqrt{\dfrac{m}{k}} [br]
I didn't think to use the acceleration equation for SHM.
Thanks for the help!
how would you derive t=2Pi square root l/g
Original post by robin.murphy1999
how would you derive t=2Pi square root l/g


This ^. I can prove the period equation for springs (the equation the OP stated) but not for a pendulum :colonhash:
so how do you get t = 2Pi (l/g)^1/2
Original post by robin.murphy1999
how would you derive t=2Pi square root l/g


Original post by TheFarmerLad
This ^. I can prove the period equation for springs (the equation the OP stated) but not for a pendulum :colonhash:


You can kind of think of it as a simple harmonic motion equation because the angular displacement of a pendulum is given as

θ=θmaxsinglT\theta=\theta_{max} \sin{\sqrt{\frac{g}{l}}T}

This is derived from Newtons laws.
Not sure how much calculus you use, but it is involved :biggrin:

Fnet=mgsinθF_{net}=-mg\sin{\theta} (from resolving the vectors when the pendulum is in motion)

For small angles θ\theta, we can use the approximation that sinθθ\sin{\theta} \approx \theta

So using the acceleration of the pendulum, we get

Fnet=mgθ=maLF_{net}=-mg\theta=maL

Using differential equation form, we get

d2θdt2+gθL=0\frac{d^2\theta}{dt^2} + \frac{g\theta}{L}=0

Solving this equation gets you to the equation at the start.

Anyway, this equation can be likened to a mass on a spring :smile:

x=AsinkmTx=A\sin{\sqrt{\frac{k}{m}}T}

As such, we can say that ω=km\omega = \sqrt{\frac{k}{m}} for springs yes?

Comparing that to the pendulum equation, we see that ω=gL\omega = \sqrt{\frac{g}{L}} and so for mass on a spring

T=2πmk=2π1ωT=2\pi\sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{1}{\omega}} so therefore the equation of a pendulum is

T=2πLgT=2\pi \sqrt{\frac{L}{g}}
As required

EDIT: Just realised you're using a=ω2xa=-\omega^2x sorry :P
Using this instead, you skip the differential equation:
Fnet=maL=mgθF_{net}=maL=mg\theta solving for aa gives a=gθLa=-\frac{g\theta}{L}

Now as θ\theta is displacement of the pendulum, θ=x\theta=x

so gxL=ω2x-\frac{gx}{L}=-\omega^2x
Therefore ω=gL\omega = \sqrt{\frac{g}{L}} so you know where to go from this (ω=2πf\omega = 2\pi f)
(edited 7 years ago)
Original post by The-Spartan
You can kind of think of it as a simple harmonic motion equation because the angular displacement of a pendulum is given as

θ=θmaxsinglT\theta=\theta_{max} \sin{\sqrt{\frac{g}{l}}T}

This is derived from Newtons laws.
Not sure how much calculus you use, but it is involved :biggrin:

Fnet=mgsinθF_{net}=-mg\sin{\theta} (from resolving the vectors when the pendulum is in motion)

For small angles θ\theta, we can use the approximation that sinθθ\sin{\theta} \approx \theta

So using the acceleration of the pendulum, we get

Fnet=mgθ=maLF_{net}=-mg\theta=maL

Using differential equation form, we get

d2θdt2+gθL=0\frac{d^2\theta}{dt^2} + \frac{g\theta}{L}=0

Solving this equation gets you to the equation at the start.

Anyway, this equation can be likened to a mass on a spring :smile:

x=AsinkmTx=A\sin{\sqrt{\frac{k}{m}}T}

As such, we can say that ω=km\omega = \sqrt{\frac{k}{m}} for springs yes?

Comparing that to the pendulum equation, we see that ω=gL\omega = \sqrt{\frac{g}{L}} and so for mass on a spring

T=2πmk=2π1ωT=2\pi\sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{1}{\omega}} so therefore the equation of a pendulum is

T=2πLgT=2\pi \sqrt{\frac{L}{g}}
As required


Brilliant, thank you
Reply 9
Show that for a body vibration simple harmonically the time period ia given by T=2π√(L/g)
Reply 10
Thankssssss
This is not correct. You get the correct answer, but you start from an incorrect equation. If you look at your F=maL equation, the units of this equation are incorrect. On the left hand side you have Netwons, and on the right hand side you have Newtons*meters. The correct way to solve this is to start with F=mgtheta, then use the arc length formula to note that theta=arclength/l. Solving for a you get a=g*arclength/l. Then using circular motion, you set a=omega*arclength. Setting the a=a you solve for omega and get omega=sqrt(g/l)

Quick Reply

Latest