Question 251: Oscillations

Question

The period of S. H.M. of a particle is 16 second. The phase difference between the positions at $\mathrm{t}=2 \mathrm{~s}$ and $\mathrm{t}=4 \mathrm{~s}$ will be

Accepted Answer

Given:
Period of SHM, $T = 16\, 	ext{seconds}$
Times: $t_1 = 2\, 	ext{s}$ and $t_2 = 4\, 	ext{s}$
Step 1:
Find the angular frequency $\omega$
The formula is:
$$ \omega = \frac{2\pi}{T} $$
Substituting $T = 16$ s:
$$ \omega = \frac{2\pi}{16} = \frac{\pi}{8}\ 	ext{rad/s} $$
Step 2:
Calculate the phase at $t_1$ and $t_2$
The general phase at time $t$ (ignoring initial phase, as only difference is required):
$$ 	ext{Phase at}\ t = \omega t $$
So,
At $t_1 = 2$ s, phase = $\omega 	imes 2 = \frac{\pi}{8} 	imes 2 = \frac{\pi}{4}$
At $t_2 = 4$ s, phase = $\omega 	imes 4 = \frac{\pi}{8} 	imes 4 = \frac{\pi}{2}$
Step 3:
Find the phase difference
$$ 	ext{Phase difference} = \left| \frac{\pi}{2} - \frac{\pi}{4} ight| = \frac{\pi}{4} $$
Final Answer:
Option C: $\boxed{\dfrac{\pi}{4}}$

Answer

$\pi$

Answer

$\frac{\pi}{2}$

Answer

$\frac{\pi}{4}$

Answer

$\frac{\pi}{8}$

Oscillations

The period of S. H.M. of a particle is 16 second. The phase difference between the positions at \mathrm{t}=2 \mathrm{~s} and \mathrm{t}=4 \mathrm{~s} will be

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