NEET-XII-Physics
12: Simple Harmonic Motion
- #11The total mechanical energy of a spring-mass system in simple harmonic motion is
E=12mω2A2.Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the same. The new mechanical energy will
(a) become 2E
(b) become E/2
(c) become
2E
(d) remain EdigAnsr: dAns : (d) remain E
Mechanical energy (E) of a spring-mass system in simple harmonic motion is given by,
`` {E}_{}=\frac{1}{2}m{\omega }^{2}{A}^{2}``
where m is mass of body, and
`` \omega `` is angular frequency.
Let m1 be the mass of the other particle and ω1 be its angular frequency.
New angular frequency ω1 is given by,
`` {\omega }_{1}=\sqrt{\frac{k}{{m}_{1}}}=\sqrt{\frac{k}{2m}}({m}_{1}=2m)``
New energy E1 is given as,
`` {E}_{1}=\frac{1}{2}{m}_{1}{\omega }_{1}^{2}{A}^{2}``
`` =\frac{1}{2}\left(2m\right)(\sqrt{\frac{k}{2m}}{)}^{2}{A}^{2}``
`` =\frac{1}{2}m{\omega }^{2}{A}^{2}=E``
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