A particle starts S.H.M. from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is 3E/4, its displacement at this instant is y = a/x. Find the value of ‘x’.

Question

A particle starts S.H.M. from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is $\frac{3 E}{4}$ , its displacement at this instant is $y = \frac{a}{x}$ . Find the value of $‘ x ’$ .

Smart Achievers · Accepted Answer

(2)

Total energy, $E = \frac{1}{2} m ω^{2} a^{2}$ ;

$K.E. = \frac{3 E}{4} = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$

So, $\frac{3}{4}$ $= \frac{a^{2} - y^{2}}{a^{2}}$ or $y^{2} = \frac{a^{2}}{4}$ or $y = \frac{a}{2}$

Solution

Q.
A particle starts S.H.M. from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is $\frac{3 E}{4}$ , its displacement at this instant is $y = \frac{a}{x}$ . Find the value of $‘ x ’$ .

Ans.
(2)

Total energy, $E = \frac{1}{2} m ω^{2} a^{2}$ ;

$K.E. = \frac{3 E}{4} = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$

So, $\frac{3}{4}$ $= \frac{a^{2} - y^{2}}{a^{2}}$ or $y^{2} = \frac{a^{2}}{4}$ or $y = \frac{a}{2}$

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Solution

Q. A particle starts S.H.M. from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is 3E4, its displacement at this instant is y=ax. Find the value of ‘x’.

Ans. (2) Total energy, E=12mω2a2; K.E.=3E4=12mω2(a2-y2) So, 34=a2-y2a2 or y2=a24 or y=a2

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Q.
A particle starts S.H.M. from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is $\frac{3 E}{4}$ , its displacement at this instant is $y = \frac{a}{x}$ . Find the value of $‘ x ’$ .

Ans.
(2)

Total energy, $E = \frac{1}{2} m ω^{2} a^{2}$ ;

$K.E. = \frac{3 E}{4} = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$

So, $\frac{3}{4}$ $= \frac{a^{2} - y^{2}}{a^{2}}$ or $y^{2} = \frac{a^{2}}{4}$ or $y = \frac{a}{2}$