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

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Answer in Brief

Answer in brief:

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

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Solution

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

L = l + r,

where l = length of string

r = radius of bob

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]

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

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