Question

Q.3
A heavy small sized sphere is suspended by
a string of length l. The sphere rotates
uniformly in a horizontal circle with the string
making an angle with the vertical. Then the
time period of this conical pendulum is-

(A) T = 2×3.14
(B) T=2×3.14×I sin theta÷g^1/2
(C) T = 2×3.14×I cos theta ÷ g^1/2
(D) T=2×3.14×I ÷ g cos theta^1/2
​

Answer 1

Answer:

Resolving T along the vertical and horizontal directions, we get Tcosθ=Mg→(i)

Tsinθ=Mμω

2

=M(lsinθ)ω

2

T=Mlω

2

Dividing equation (ii) by (i)

We get

cosθ

1

=

g

lω

2

ω

2

=

lcosθ

g

∴ time period

t=

ω

2pi

=2π

g

lcosθ