A particle doing S.H.M. have amplitude 5 cm, mass 0.5 kg and angular frequency 5 is at 1 cm from mean position. Find potential energy and kinetic energy.

Question

A particle doing S.H.M. have amplitude 5 cm, mass 0.5 kg and angular frequency 5 is at 1 cm from mean position. Find potential energy and kinetic energy. (A) K.E. =150 × 10-4 J, P.E. =6.25 × 10-4 J (B) K.E. =6.25 × 10-4 J , P . E .=150 × 10-3 J (C) K.E. =6.25 × 10-4 J, P.E. =6.25 × 10-4 J (D) K.E. =150 × 10-3 J, P.E. =150 × 10-3 J

Tardigrade Admin · Accepted Answer

Potential energy U (1/2)m 2x2 (1/2) 0.5 25 (102)2 =(25/4)×10-4 Joule = 6.25 Ã 10-4 Joule Kinetic energy K (1/2)m 2(A2 x2) =(1/2)×0.5×25[25-1]×10-4 = 150 Ã 10-4 joule

Q. A particle doing S.H.M. have amplitude $5 c m$ , mass $0.5 k g$ and angular frequency $5$ is at $1 c m$ from mean position. Find potential energy and kinetic energy.

K.E. $= 150 \times 1 0^{- 4} J$ , P.E. $= 6.25 \times 1 0^{- 4} J$

K.E. $= 6.25 \times 1 0^{- 4} J, P . E . = 150 \times 1 0^{- 3} J$

K.E. $= 6.25 \times 1 0^{- 4} J$ , P.E. $= 6.25 \times 1 0^{- 4} J$

K.E. $= 150 \times 1 0^{- 3}$ J, P.E. $= 150 \times 1 0^{- 3} J$

Potential energy
U $\frac{1}{2} m^{2} x^{2} \frac{1}{2} 0.5 25 (1 0^{2})^{2}$
$= \frac{25}{4} \times 1 0^{- 4}$ Joule = 6.25 × 10^-4 Joule
Kinetic energy
$K \frac{1}{2} m^{2} (A^{2} x^{2})$
$= \frac{1}{2} \times 0.5 \times 25 [25 - 1] \times 1 0^{- 4}$
= 150 × 10^-4 joule

Q. A particle doing S.H.M. have amplitude 5cm, mass 0.5kg and angular frequency 5 is at 1cm from mean position. Find potential energy and kinetic energy.

K.E. =150×10−4J, P.E. =6.25×10−4J

K.E. =6.25×10−4J,P.E.=150×10−3J

K.E. =6.25×10−4J, P.E. =6.25×10−4J

K.E. =150×10−3 J, P.E. =150×10−3J