A pendulum completes 2 oscillations in 5s. (a) What is its time period? (b) If g = 9.8 ms-2, find its length.

2 oscillations in 5 seconds, so

frequency per second = 2/5

⇒ frequency per second = 0.4 Hz

Hence, frequency of oscillation = 0.4 hertz.

As we know,

Time Period (T) = 1/f

So, substituting the value of f = 0.4 hertz, in the equation above we get,

Time Period (T) = 1/0.4

⇒ T = 2.5 s

Hence, time period of the pendulum is 2.5 s

(b) As we know,

T = 2 \pi \sqrt{l\over g}

Given,

g = 9.8 ms^-2

and we know,

π = 3.14

T = 2.5 s

Substituting the values in the formula above we get,

2.5 = 2 × 3.14 × \sqrt {l\over9.8}

⇒ ({2.5\over2 \times 3.14})={\sqrt {l\over9.8}}

⇒ ({2.5\over 6.28})^2={l\over9.8}

⇒ (0.398)^2={l\over9.8}

⇒ 0.158={l\over9.8}

⇒ l = 9.8 × 0.158 = 1.55

Hence, the length of a seconds’ pendulum = 1.55 m