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

i) T cosθ cosθ = Mg.

ii) T + Mg cosθ = Mv²/L.

iii) The magnitude of tangential acceleration of the bob
${\sf{{(|a_{T}}}}{\sf{{|= g \: sinθ_{}}}}$
iv) T = Mg (3cosθ – 2cos∅)​

Answer 1

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \red{ \underline{ \sf \: from \: the \: figure : }} \\ \\ \hookrightarrow \: \bf \:Net \: force \: towards \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \: the \: centre \: , \:F_N =T - Mg \: cos \theta \\ \\ \\ \hookrightarrow\:\bf \: Centrifugal \: force \: , \: F_C= \frac{M {v}^{2} }{L} \\ \\ \\ \pink{ \underline{ \text{for \: the \: pendulum}}} \\ \\ \bf \: F_N = F_C \\ \\ \implies \blue{ \boxed{{ \orange{\bf \: T - Mg \: cos \theta = \frac{M{v}^{2} }{L} }}}} \\ \\ \end{array}}}}}$

So, correct condition is :

$\green{ \boxed{ \begin{array}{cc} \bf \: T - Mg \: cos \theta = \frac{M {v}^{2} }{L} \end{array}}}$

Answer 2

${\red{ \boxed{ \boxed{ \begin{array}{cc} \red{ \underline{ \sf \: From \: the \: figure : }} \\ \\ \hookrightarrow \: \bf \:Net \: force \: towards \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \: the \: centre \: , \:F_N =T - Mg \: cos \theta \\ \\ \\ \hookrightarrow\:\bf \: Centrifugal \: force \: , \: F_C= \frac{M {v}^{2} }{L} \\ \\ \\ \pink{ \underline{ \text{For \: the \: pendulum}}} \\ \\ \bf \: F_N = F_C \\ \\ \implies \blue{ \boxed{{ \orange{\bf \: T - Mg \: cos \theta = \frac{M{v}^{2} }{L} }}}} \\ \\ \end{array}}}}}$

So, correct condition is :

$\green{ \boxed{ \begin{array}{cc} \tt \: T - Mg \: cos \theta = \frac{M {v}^{2} }{L} \end{array}}}$