Question

Q10. A pendulum completes 2 oscillations in 5 seconds.



if g is equal to 9.8 metre per second square, find its length. ​

Answer 1

ANSWER

We know that time period of a simple pendulum

T=2π√l/g

squaring on both sides , we get

T^2= 4π^2(l/g)

from here

l =(T^2×g)/4π^2

given that

T=5s for two oscillations

so, for one oscillation T=5/2=2.5s

g=9.8 m/s^2

so,

l = (5^2×9.8)/4×3.14^2

=6.2mts

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Answer 2

Given that, a pendulum completes 2 oscillations in 5 seconds. If g is 9.8 m/s².

Here we have; T = 5 sec and g = 9.8 m/s².

We have to find the length of the pendulum.

Now,

T = 2π √(l/g)

Squaring both sides,

(T)² = [ 2π √(l/g) ]²

On solving we get,

T² = 4π² l/g

• For two oscillations:

Time period = 5 sec

• For one oscillation:

Time period = (Time for two oscillations)/2

= 5/2 = 2.5 sec

Substitute the Known values,

→ (2.5)² = 4 × 3.14 × 3.14 × l/9.8

→ 6.25 = 39.44 × l/9.8

→ 6.25 = 4.02 × l

→ 6.25/4.02 = l

→ 1.55 = l

→ l = 1.55 m

Therefore, the length of the pendulum is 1.55 m.