A body is executing SHM under the action of a force whose maximum magnitude is $50 N$ . The magnitude of force acting on the particle at the time, when its energy is half kinetic and half potential is

25 N

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25 \sqrt{2} N

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\frac{25}{\sqrt{2}} N

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12.5 N

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Solution

The correct option is B $25 \sqrt{2} N$
$K E = P E$
$\frac{1}{2} m w^{2} (A^{2} - x^{2}) = \frac{1}{2} m w^{2} x^{2}$
$A^{2} = 2 x^{2}$
$x = A / \sqrt{2}$
Force $= m w^{2} x$
$= \frac{m w^{2} A}{\sqrt{2}}$
$= \frac{50}{\sqrt{2}}$
$= 25 \sqrt{2} N$

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A body is executing SHM under the action of a force whose maximum magnitude is 50N. The magnitude of force acting on the particle at the time, when its energy is half kinetic and half potential is

The correct option is B 25√2NKE=PE12mw2(A2−x2)=12mw2x2A2=2x2x=A/√2Force =mw2x =mw2A√2 =50√2 =25√2N

A body is executing SHM under the action of a force whose maximum magnitude is $50 N$ . The magnitude of force acting on the particle at the time, when its energy is half kinetic and half potential is

The correct option is B $25 \sqrt{2} N$
$K E = P E$
$\frac{1}{2} m w^{2} (A^{2} - x^{2}) = \frac{1}{2} m w^{2} x^{2}$
$A^{2} = 2 x^{2}$
$x = A / \sqrt{2}$
Force $= m w^{2} x$
$= \frac{m w^{2} A}{\sqrt{2}}$
$= \frac{50}{\sqrt{2}}$
$= 25 \sqrt{2} N$