A simple pendulum of length L and mass M is oscillating in a plane about a vertical line between angular limits $- ϕ$ and $+ ϕ$ . For an angular displacement $θ (| θ | < ϕ)$ the tension in the string and velocity of the bob are T and v respectively. The following relations hold good under the above condition.

T c o s θ = M g

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T + M g c o s θ = \frac{M v^{2}}{L}

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The magnitude of tangential acceleration of the bob

| a_{T} | = g s i n θ

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T = M g c o s ϕ

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Solution

The correct options are
C The magnitude of tangential acceleration of the bob $| a_{T} | = g s i n θ$
D $T = M g c o s ϕ$

$T - m g c o s θ = \frac{m v^{2}}{L}$
and $| a_{T} | = g s i n θ$
At maximum angular displacement
$T - m g c o s ϕ = 0$
so $T = m g c o s ϕ$
Hence, (C), (D)

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Similar questions

Q. A simple pendulum of length

L

and mass (bob)

M

is oscillating in a plane about a vertical line between angular limit

- ϕ

and

+ ϕ

. For an angular displacement

θ (| θ | < ϕ)

, the tension in the string and the velocity of the bob are

T

and

V

respectively. The following relations hold good under the above conditions:

A simple pendulum of length L and mass (bob) M is oscillating in a plane about a vertical line between angular limits -∅and +∅. For an angular displacement $θ (| θ | < ϕ)$ , the tension in the string and the velocity of the bob are T and v respectively. The following relations hold good under the above conditions