The total energy of a particle executing SHM is 80 J. What is the potential energy when the particle is ata distance of 3/4 of amplitude from the mean position?

Question

The total energy of a particle executing SHM is 80 J. What is the potential energy when the particle is ata distance of 3/4 of amplitude from the mean position? (A) 60 J (B) 10 J (C) 40 J (D) 45 J

Tardigrade Admin · Accepted Answer

Given, (1/2) m ω2 r2=80 J ; ∴ P E=(1/2) m ω2 y2=(1/2) m ω2 ×((3/4) r)2 =(9/16)((1/2) m ω2 r2)=(9/16) × 80=45 J

Q. The total energy of a particle executing SHM is $80 J$ . What is the potential energy when the particle is ata distance of $3/4$ of amplitude from the mean position?

60 J

10 J

40 J

45 J

Given, $\frac{1}{2} m ω^{2} r^{2} = 80 J;$
$∴ PE = \frac{1}{2} m ω^{2} y^{2} = \frac{1}{2} m ω^{2} \times (\frac{3}{4} r)^{2}$
$= \frac{9}{16} (\frac{1}{2} m ω^{2} r^{2}) = \frac{9}{16} \times 80 = 45 J$

Q. The total energy of a particle executing SHM is 80J. What is the potential energy when the particle is ata distance of 3/4 of amplitude from the mean position?