Question

A pendulum has a length 1 m and a mass 1 kg. Assume Earth free fall acceleration equal to 10 m/s^2. When the pendulum oscillates, the maximal deflection angle is +/- 1 degree.

a) If the maximal deflection angle doubles, what would be the new period?

b) If the pendulum mass goes up 100 times, what would be the new period?

c) If the length goes up 100 times, what would be the period?

d) If the pendulum moves to Moon, what would be the period?

Answer 1

(a) If the maximal deflection angle doubles

The relation between time (T ) and angle of deflection (θ) can be derived by solving the equation of motion of the simple pendulum (from F=ma)

−gsinθ=lθ¨ (here θ¨ is double derivative of θ - that angluar acceleartion)

For small angles, θ≪1and hence sinθ≈θ. Hence,

θ¨=−(g/l) x θ

Solution forthis second-order differential equation is

θ = θ₀cos(ωt), here ω=√(g/l)

The period is thus T=2πω=2π√(l/g), which is independent of the starting angle θ₀

For large angles, however, the above derivation is invalid.

Without going into the derivation, the general expression of period is

At large angles, the {θ₀²/16} term starts to grow big and cause T to deviate from T= 2π√(l/g)

Using this to solve gives  

Initial time period is (for θ₀=1) ,

T= 2π√(1/10) x (1+1²/16) ......

T = 2.11 sec

New time period is (for θ₀=2) ,

T= 2π√(1/10) x (1+2²/16) ......

T = 2.48 sec

----------------------------------------------------------------------

(b) If the pendulum mass goes up 100 times , there will be no effect on Time period as equation T= 2π√(l/g) is independent of mass.

Ans - No effect on Time period

----------------------------------------------------------------------

(c)  If the length goes up 100 times, put in values in   T= 2π√(l/g) gives  

T= 2π√(100 l/g)

T= 10 x 2π√(l/g) ...... New time period

Ans - New time period will be 10 times more than initial time period.

----------------------------------------------------------------------

(d ) If the pendulum moves to Moon, assuming gravity of moon is 1/6th of earth ,

g_moon= g_earth/ 6

put in values in   T= 2π√(l/g) gives

T= 2π√ ( l /g_moon)

T= 2π√ ( l /(g_earth/ 6))

T= √6 x 2π√( l /g_earth) ...... New time period

Ans - time period for moon will be √6 times more than time period of earth .