Question

compare the time period of two pendulum of length 2m and 18m (g=9.8metre per square second)

Answer 1

Given :

▪ Length of pendulum a = 2m

▪ Length of pendulum b = 18m

▪ Acc. due to gravity = 9.8m/s²

To Find :

▪ Relation b/w time period of both pendulums.

Solution :

☞ Formula of time period of a simple pendulum in terms of length of pendulum and acc. due to gravity is given by

• T = 2π √(l/g)

where,

T denotes time period

l denotes length of simple pendulum

g denotes acc. due to gravity

Pendulum - a :

→ Ta = 2π √(la/g)

→ Ta = 2π √(2/g) .......... (i)

Pendulum - b :

→ Tb = 2π √(lb/g)

→ Tb = 2π √(18/g) .......... (ii)

Taking (i) and (ii), we get

↗ Ta/Tb = √2/√18

↗ Ta/Tb = √1/√9

↗ Ta/Tb = 1/3

↗ Ta : Tb = 1 : 3

Answer 2

Answer:

YOUR ANSWER :

Compare the time period of two pendulum of length 2m and 18m (g=9.8metre per square second)

YOUR ANSWER :

Given :

• Length of pendulum a = 2m
• Length of pendulum b = 18m
• Acc. due to gravity = 9.8m/s²

To Find :

• Relation b/w time period of both pendulums.

Solution :

✔ Formula of time period of a simple pendulum in terms of length of pendulum and acc.

• T = 2π √(l/g)

Pendulum - a :

=> Ta = 2π √(la/g)

=> Ta = 2π √(2/g) ....... (i)

Pendulum - b :

=> Tb = 2π √(lb/g)

=> Tb = 2π √(18/g) ....... (ii)

Taking (i) and (ii), We get

Ta/Tb = √2/√18

=> Ta/Tb = √1/√9

=> Ta/Tb = 1/3

=> Ta : Tb = 1 : 3

ANSWER : 1 : 3.