Question

proof expression of time period for simple pendulum ?​

Answer 1

$\huge\bf{\pink{\mid{\underline{\overline{your\:answer}}}\mid}}$

$\star{\mathtt{\pink{\underline{Simple \: pendulum-}}}}$ If a heavy point mass is suspended by a weightless, inextensible and perfectly fexible string from a rigid support, then this arrangement is called a Simple pendulum.

$\large{\underline{\underline{\mathbf{Expression\: for\: time\: period}}}}$

★ $\large\bf{\underline{Method-I \: (force \: method) }}$

For small angular displacement,

sin∅ ≈ ∅

so that,

F = -mg sin∅

=> -mg∅

then,

$= > \: \: - ( \frac{ mg}{l} )y$

$= > \: \: - ky$

$= > \: \: k = \frac{mg}{l}$

( y = l ∅)

$\bold{thus, \: the\: time\: period\: of\: simple\: pendulum\: is, }$

$= > \: t = 2\pi \sqrt{ \frac{m}{k} }$

$\large\boxed{t=2\pi\sqrt{\frac{l}{g}}}$

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★ $\large\bf{\underline{Method-II \: (Torque\: method) }}$

Bob of pendulum moves along the arc of circle in vertical plane. motion involved is angular and oscillatory, where restoring torque is provided by gravitational force.

t = -(mg)( l sin ∅)

$\rm{\underline{negative\:sign\:show\:opposite\:direction\:of\:torque}}$

t = -mg l∅

If angular displacement is small,then

Sin∅ ≈ ∅

I = ( moment of inertia of bob)

$i \alpha = - mgl$

Now,

(-mgl / I)

$\bold{compairing\:with\:standard\:diffrential\:equation\:of\:angular\:SHM}$

therefore,

$\large\boxed{t=2\pi\sqrt{\frac{l}{g}}}$

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Answer 2

$\huge{\mathfrak{\underline{\underline{Solution:-}}}}$

$\sf{Let\:t\propto m^{a}\:l^{b}\:g^{c}}$

Where a, b and c are the power of m (mass), l (length) and g (acceleration due to gravity).

$\sf{or\:t=k\:m^{a}\:l^{b}\:g^{c}\:\:\:\:\:.......(1)}$

Where k is dimension less constant of proportionality.

Where k is dimension less constant of proportionality.Writing the dimensions in the terms of M, L and T on each side of equation (1), we get..

$\sf{\big[M^{0}\:L^{0}\:T^{1}\big]=M^{a}\:L^{b}\:\big(LT^{-2}\big)^{c}}$

$\sf{=M^{a}\:L^{b+c}\:T^{-2c}}$

Applying the principal of homogeneity of dimension, we get..

$\sf{a=0\:\:\:\:\:..........(2)}$

$\sf{b+c=0\:\:\:\:\:..........(3)}$

$\sf{or,\:c=\frac{-1}{2}\:\:\:\:\:..........(4)}$

Putting the value of c in equation (3), we get

$\sf{b+\bigg(\frac{-1}{2}\bigg)=0}$

$\sf{or,\:b=\frac{1}{2}}$

Now putting the value of a, b and c in equation (1), we get

$\sf{t=k\:m^{0}\:l^{1/2}\:g^{-1/2}}$

$\sf{or,\:t=k\sqrt{\frac{l}{g}}}$

Using other method, we calculate the value of k =

$2\pi$

$\sf{\boxed{\boxed{So,\:t=2\pi \sqrt{ \frac{l}{g}}}}}$