Correct Answer is Option (C)

(a)

Amplitude of oscillation is continuously decreasing. It means bob of pendulum oscillate with air friction.

(b)

$F=-kx$ (restoring force of a spring)

(c)

Amplitude of oscillation is remains same. It means bob of pendulum is oscillating under negligible air resistance.

(d)

$P.E.=\frac{1}{2}k{x}^2,K.E.=\frac{1}{2}m{\omega }^2{A}^2-\frac{1}{2}k{x}^2$

$T.E.=\frac{1}{2}m{\omega }^2{A}^2=$ constant

Correct Answer is Option (C)

Displacement equation of SHM of frequency 'n'

x = A sin ($\omega$t) = A sin (2$\pi$nt)

Now,

Potential energy

$U=\frac{1}{2}k{x}^2=\frac{1}{2}K{A}^2{\sin }^2(2\pi nt)$

$=\frac{1}{2}k{A}^2\left[\frac{1-\cos (2\pi (2n)t)}{2}\right]$

So frequency of potential energy = 2n

Correct Answer is Option (C)

Kinetic energy + potential energy = total energy
When kinetic energy is maximum, potential energy is zero and vice versa.

$\therefore$ Maximum potential energy = total energy.

0 + K₀ = K₀ (K.E. + P.E. = total energy).

Correct Answer is Option (C)

Potential energy of simple harmonic oscillator = $\frac{1}{2}m{\omega }^2{y}^2$

for $y=\frac{a}{2},P.E=\frac{1}{2}m{\omega }^2\frac{a^2}{4}$

$\Rightarrow P.E=\frac{1}{4}\left(\frac{1}{2}m{\omega }^2{a}^2\right)=\frac{E}{4}$

Correct Answer is Option (A)

Potential energy of particle performing SHM varies parabolically in such a way that at mean position it becomes zero and maximum at extreme position.

Correct Answer is Option (C)

For a simple harmonic motion between A and B, with O as the mean position, maximum kinetic energy of the particle executing SHM will be at O and maximum potential energy will be at A and B.


$\therefore$ Displacement between maximum potential energy and maximum kinetic energy is $\pm a$.

Correct Answer is Option (B)

Energy = $\frac{1}{2}m{\omega }^2{a}^2=\frac{1}{2}k{a}^2$