
In a simple harmonic oscillator, the total mechanical energy (T.E.) is given by:
\[ T.E. = \frac{1}{2}kA^2, \] where \( k \) is the spring constant and \( A \) is the amplitude of oscillation.
- Since the amplitude \( A \) remains the same, the total mechanical energy (T.E.) will also remain the same, as it depends only on \( k \) and \( A \), not on the mass \( m \) of the oscillating particle.
Thus, even if the mass of \( P \) is doubled, the total mechanical energy \( E \) will remain unchanged.
Answer: E
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)