CGPET 2021 Solved Paper

(a) We know that, centripital acceleration,
gc=

v2

a

For circular path, effective acceleration of pendulum,
geff2=g2+gc2
⇒ geff=√g2+gc2
⇒ geff=√g2+(

v2

a

)2

Now, time period, T=2π√

l

geff

When the railway carraiage is travelling, then frequency of oscillation of simple pendulum is given as

1

n

=2π√

l

geff

.....(i)(∵T=

1

n

)
and when v=0 (stationary state), then

1

n1

=2π√

l

g

.....(ii)
On dividing Eq. (i) by Eq. (ii), we get

1 n

1 n1

=

2π√ l geff

2π√ l g

⇒

n1

n

=√

g

geff

⇒

n12

n2

=

g

geff

⇒

n12

n2

=

g

√g2+( v2 a )2

Squaring both side of the above equation, we get
(

n1

n

)4=

g4

g2+( v2 a )2

⇒ g2+(

v2

a

)2=(

n

n1

)4g2
⇒ (v2)2=a2g2[(

n

n1

)4−1]
⇒ v2=ag‌√(

n4

n14

−1)