Question

The time period of oscillation of a thin
uniform rod when suspended from a point P
and when suspended from the point Q, P
and Q being on opposite sides of the centre
of mass, is same and is equal to T. If the
distance between P and Q is I, the
acceleration due to gravity g=​

Answer 1

Answer:

Explanation:

The answer is 4π^2l/T^2

Answer 2

The acceleration due to gravity is given by

$\boxed{g=\frac{4\pi^2l}{gT^2}}$

Explanation:

We know that

Time period of a physical pendulum is given by

$T=2\pi\sqrt{\frac{I}{MgL_{cm}}}$

Where I is the moment of inertia about the support

M is the mass of the pendulum

$L_{cm}$ is the distance of the centre of mass from the support

Therefore,

For the given question

Let the mass of the rod is M and total length L

Moment of inertia of a uniform rod of mass M and length L about its centre of mass is given by

$I=\frac{1}{12}ML^2$

When the rod is suspended from point P, if the distance of P from the centre of mass of the rod is x then

$T=2\pi\sqrt{\frac{(1/12)ML^2+Mx^2}{Mgx}}$

or, $T^2(Mgx)=4\pi^2[(1/12)ML^2+Mx^2]$

or, $T^2gx=4\pi^2[(1/12)L^2+x^2]$        ..... (1)

When the rod is suspended from Q

$T=2\pi\sqrt{\frac{(1/12)ML^2+M(l-x)^2}{Mg(l-x)}}$

or, $T^2g(l-x)=4\pi^2[(1/12)L^2+(l-x)^2]$  ....... (2)

Subtracting eq (1) from eq (2)

$T^2g(l-2x)=4\pi^2[(l-x)^2-x^2]$

$\implies T^2g(l-2x)=4\pi^2(l-2x)(l)$

$\implies T^2g=4\pi^2l$

$\implies g=\frac{4\pi^2l}{gT^2}$

Hope this answer is helpful.

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