SATHEE: Oscillations - Result Question 24

Oscillations - Result Question 24

27. The particle executing simple harmonic motion has a kinetic energy $K_{0} \cos^{2} ω t$ . The maximum values of the potential energy and the total energy are respectively

(a) $K_{0} / 2$ and $K_{0}$

(b) $K_{0}$ and $2 K_{0}$

(d) 0 and $2 K_{0}$ .

[2007]

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Answer:

Correct Answer: 27. (c)

Solution:

Also, we know that, in S.H.M., when potential energy is maximum, K.E. is zero and vice-versa.

$∴ U_{max} + 0 = E \Rightarrow U_{max} = E$

Further,

$K . E . = \frac{1}{2} m ω^{2} a^{2} \cos^{2} ω t$

But by question, $K . E . = K_{0} \cos^{2} ω t$

$∴ K_{0} = \frac{1}{2} m ω^{2} a^{2}$

Hence, total energy, $E = \frac{1}{2} m ω^{2} a^{2} = K_{0}$

$Misplaced &$ .

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Oscillations - Result Question 24

27. The particle executing simple harmonic motion has a kinetic energy K0cos2⁡ωt. The maximum values of the potential energy and the total energy are respectively

Answer:

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27. The particle executing simple harmonic motion has a kinetic energy $K_{0} \cos^{2} ω t$ . The maximum values of the potential energy and the total energy are respectively