# Derive an Expression for the Period of Motion of a Simple Pendulum. On Which Factors Does It Depend? - Physics

Derive an expression for the period of motion of a simple pendulum. On which factors does it depend?

#### Solution

a) Consider a simple pendulum of mass ‘m’ and length ‘L’.

L = l + r,

where l = length of string

b) Let OA be the initial position of pendulum and OB, its instantaneous position when the
the string makes an angle θ with the vertical.

In the displaced position, two forces are acting on the bob:

• Gravitational force (weight) ‘mg’ in a downward direction
• Tension T′ in the string.

c) Weight ‘mg’ can be resolved into two rectangular components:

• Radial component mg cos θ along OB and
• Tangential component mg sin θ perpendicular to OB and directed towards mean
position.

d) mg cos θ is balanced by tension T′ in the string, while mg sin θ provides restoring force

∴ F = − mg sin θ

where a negative sign shows that force and angular displacement are oppositely directed.
Hence, restoring force is proportional to sin θ instead of θ. So, the resulting motion is not
S.H.M.

e) If θ is very small then,

sin θ ≈ θ = "x"/"L"

∴ F = - "mg" "x"/"L"

∴ "F"/"m" = - "g" "x"/"L"

∴ "ma"/"m" = - "g""x"/"L"

∴ "a" = - "g"/"L" "x"  ....(i)

∴ a α - x    ....[therefore "g"/"L" = "constant"]

f) In S.H.M,

a = − ω2 x ….(ii)

Comparing equations (i) and (ii), we get,

ω^2 = "g"/"L"

But, ω = (2pi)/"T"

therefore ((2pi)/"T")^2 = "g"/"L"

therefore (2pi)/"T" = sqrt("g"/"L")

therefore "T" = 2pi sqrt("L"/"g")   ....(iii)

Equation (iii) represents time period of simple pendulum.

g) Thus period of a simple pendulum depends on the length of the pendulum and
acceleration due to gravity.

Is there an error in this question or solution?
Chapter 5: Oscillations - Exercises [Page 129]

#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 5 Oscillations
Exercises | Q 7 | Page 129

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