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प्रश्न
The total mechanical energy of a spring-mass system in simple harmonic motion is \[E = \frac{1}{2}m \omega^2 A^2 .\] 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
पर्याय
become 2E
become E/2
become \[\sqrt{2}E\]
remain E
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उत्तर
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,
\[E_1 = \frac{1}{2} m_1 \omega_1^2 A^2 \]
\[ = \frac{1}{2}(2m)(\sqrt{\frac{k}{2m}} )^2 A^2 \]
\[ = \frac{1}{2}m \omega^2 A^2 = E\]
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