MHT CET 2024 10th May Evening Shift | Simple Harmonic Motion Question 18 | Physics

A simple pendulum of length ' $l$ ' has a brass bob attached at its lower end. It's period is ' T '. A steel bob of the same size, having density ' $x$ ' times that of brass, replaces the brass bob. Its length is then so changed that the period becomes ' 2 T '. What is the new length?

$4 l x$

$\frac{4 l}{\mathrm{x}}$

$4 l$

$2 l$

MHT CET 2024 10th May Morning Shift

MCQ (Single Correct Answer)

-0

A particle performing S.H.M. with maximum velocity ' $V$ '. If the amplitude double and periodic time is made, $\left(\frac{1}{3}\right)^{\text {rd }}$ then the maximum velocity is

$\frac{\mathrm{V}}{3}$

$\frac{3 V}{2}$

$\frac{2 \mathrm{~V}}{3}$

MHT CET 2024 10th May Morning Shift

MCQ (Single Correct Answer)

-0

Let ' $l_1$ ' be the length of simple pendulum. Its length changes to ' $l_2$ ' to increase the periodic time by $20 \%$. The ratio $\frac{l_2}{l_1}=$

1.22

1.33

1.44

1.55

MHT CET 2024 9th May Evening Shift

MCQ (Single Correct Answer)

-0

A particle is executing a linear simple harmonic motion. Let ' $\mathrm{V}_1$ ' and ' $\mathrm{V}_2$ ' are its speed at distance ' $x_1$ ' and ' $x_2$ ' from the equilibrium position. The amplitude of oscillation is

$\sqrt{\frac{V_1^2 x_2^2-V_2^2 x_2^2}{V_1^2-V_2^2}}$

$\sqrt{\frac{V_1^2-V_2^2}{V_1^2 x_2^2-V_2^2 x_1^2}}$

$\sqrt{\frac{V_1^2 x_2^2-V_2^2 x_1^2}{V_1^2-V_2^2}}$

$\sqrt{\frac{V_1^2 x_1^2-V_2^2 x_2^2}{V_1^2-V_2^2}}$

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