Question

Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. What is seconds pendulum?

Answer 1

A2A.

First of all take note of the fact that for small values of x, sinx = x.

Now picture the pendulum in your mind. The bob is subject to both tension through the string and gravitational force down the earth.

Now as the bob attains potential energy in the uniform gravitational field note the angular displacement with respect to its mean position. Let it be x.

If T is the tension in string and m be the mass of the bob

T=mgcosx

Now the force left to produce torque about the point of suspension is

mgsinx = F

Angle between F and l is -x

Torque therefore is

τ = -mglsinx (where l is the length of the pendulum)

If x ≈ 0 then sinx ≈ x

τ = -mglx

αml² = -mglx

Therefore , α = -g/l(x)

Interpretation of above equation : angular acceleration towards mean position increases as x increases. Therefore for small amplitude the motion of simple pendulum is simple harmonic.

Also since , g/l = ω²

ω = √g/l

Time period, T = 2π/ω = 2π√l/g

Answer 2

Hey mate,

● Motion of pendulum as SHM -
[Refer to the figure]

Consider a simple pendulum with mass m and length l.
Let y and θ be its linear and angular displacements respectively.
θ = y/l

At extreme position,
Tangential acceleration is-
a = gsinθ
Torque about point of suspension is-
τ = -Fl
τ = -l(mgsinθ)
M.I. is related by-
τ = Iα
-l(mgsinθ) = Iα
When, θ is small, sinθ = θ,
α = -lmgθ/I

Comparing this with std eqn of SHM ω = √(mgl/I),
We can say that pendulum shows SHM.

● Period of oscillation-
T = 2π/ω
T = 2π/√(mgl/I)
Here, I = ml^2
T = 2π/√(mgl/ml^2)
T = 2π√(l/g)

Hope that was useful...