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PHXI14:OSCILLATIONS

364146 A small body of mass 10 gram is making harmonic oscillations along a straight line with a time period of $π / 4$ and the maximum displacement is $10 c m$ . The energy of oscillation is

0.32 \times 10^{- 2} J

0.16 \times 10^{- 2} J

0.48 \times 10^{- 2} J

0.56 \times 10^{- 2} J

Explanation:

$E = \frac{1}{2} m ω^{2} a^{2}$
$= \frac{1}{2} \times \frac{10}{1000} \times 64 \times 10^{- 2}$
$= 32 \times 10^{- 4} J$

PHXI14:OSCILLATIONS

364147 The total energy of the body executing simple harmonic motion is $E$ then, the kinetic energy when the displacement is at the mid point of mean and extreme position is

\frac{E}{4}

\frac{E}{2}

\frac{\sqrt{3} E}{4}

\frac{3 E}{4}

Explanation:

Total energy of SHM,
$E = \frac{1}{2} m ω^{2} a^{2}, (where, a = amplitude)$
Kinetic energy,
$K = \frac{1}{2} m ω^{2} (a^{2} - y^{2}) \Rightarrow K = E - \frac{1}{2} m ω^{2} y^{2}$
When $y = \frac{a}{2} \Rightarrow K = E - \frac{1}{2} m ω^{2} (\frac{a^{2}}{4}) = E - \frac{E}{4}$
$K = \frac{3 E}{4}$

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PHXI14:OSCILLATIONS

364146 A small body of mass 10 gram is making harmonic oscillations along a straight line with a time period of $π / 4$ and the maximum displacement is $10 c m$ . The energy of oscillation is

0.32 \times 10^{- 2} J

0.16 \times 10^{- 2} J

0.48 \times 10^{- 2} J

0.56 \times 10^{- 2} J

Explanation:

$E = \frac{1}{2} m ω^{2} a^{2}$
$= \frac{1}{2} \times \frac{10}{1000} \times 64 \times 10^{- 2}$
$= 32 \times 10^{- 4} J$

PHXI14:OSCILLATIONS

364147 The total energy of the body executing simple harmonic motion is $E$ then, the kinetic energy when the displacement is at the mid point of mean and extreme position is

\frac{E}{4}

\frac{E}{2}

\frac{\sqrt{3} E}{4}

\frac{3 E}{4}

Explanation:

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