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Standard XII

Physics

SHM expression

The total ene...

Question

The total energy of particle executive SHM is $80 J$ . What is the kinetic energy of the particle when it is at a distance of $\frac{3}{4}$ of amplitude from the mean positive?

45 J

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35 J

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55 J

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25 J

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Solution

The correct option is D $35 J$
Total energy of particle in SHM, $U = \frac{1}{2} m ω^{2} a^{2}$

$\frac{U}{E} = \frac{\frac{1}{2} m ω^{2} y^{2}}{\frac{1}{2} m ω^{2} a^{2}} = \frac{y^{2}}{a^{2}} \Rightarrow \frac{U}{80} = \frac{{(\frac{3}{4} a)}^{2}}{a^{2}} = \frac{9}{16} \Rightarrow U = 45 J$

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The total energy of particle executive SHM is 80J. What is the kinetic energy of the particle when it is at a distance of 34 of amplitude from the mean positive?

The correct option is D 35JTotal energy of particle in SHM, U=12mω2a2UE=12mω2y212mω2a2=y2a2⇒U80=(34a)2a2=916⇒U=45 J

The total energy of particle executive SHM is $80 J$ . What is the kinetic energy of the particle when it is at a distance of $\frac{3}{4}$ of amplitude from the mean positive?

The correct option is D $35 J$
Total energy of particle in SHM, $U = \frac{1}{2} m ω^{2} a^{2}$

$\frac{U}{E} = \frac{\frac{1}{2} m ω^{2} y^{2}}{\frac{1}{2} m ω^{2} a^{2}} = \frac{y^{2}}{a^{2}} \Rightarrow \frac{U}{80} = \frac{{(\frac{3}{4} a)}^{2}}{a^{2}} = \frac{9}{16} \Rightarrow U = 45 J$