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Hint: The oscillation frequency (f) is measured in hertz (Hz), the number of oscillations per second. The time is called the period (T) for one oscillation and is measured in seconds. Calculate Inertia, center of mass and put it in the formula of time period.
Formula used:
$\mathrm{T}=2 \pi \sqrt{\dfrac{\mathrm{I}}{\mathrm{mg} \mathrm{y}_{\mathrm{cm}}}}$
T = Time period
m = mass of the body
g= gravitational acceleration
y = Center of the mass (Y coordinate)
Complete Step-by-Step solution:
Objects can oscillate in all kinds of ways, but SHM and Simple Harmonic Motion are a really important form of oscillation. The object's acceleration is directly proportional to its displacement from its position in equilibrium.
The frequency (f) of an oscillation is measured in hertz (Hz) it is the number of oscillations per second. The time for one oscillation is called the period (T) it is measured in seconds.
$f=\dfrac{1}{T}$
$\mathrm{T}=2 \pi \sqrt{\dfrac{\mathrm{I}}{\mathrm{mg} \mathrm{y}_{\mathrm{cm}}}}$
$\mathrm{I}=\dfrac{\mathrm{MR}^{2}}{2}-\left(\dfrac{1}{2} \dfrac{\mathrm{M}}{4}\left(\dfrac{\mathrm{R}}{2}\right)^{2}+\dfrac{\mathrm{M}}{4}\left(\dfrac{\mathrm{R}}{2}\right)^{2}\right)$
$=\dfrac{13 \mathrm{MR}^{2}}{32} \mathrm{y}_{\mathrm{cm}}=\dfrac{\mathrm{R}}{6}$
$\therefore \text{T}=2\pi \sqrt{\dfrac{13}{4}\quad \dfrac{\text{R}}{\text{g}}}$
The time period of small oscillations of remaining portion about $O$ is $\pi \sqrt{\dfrac{13R}{g}}$
Hence, the correct option is (B).
Note:
In simple harmonic motion, repetitive back and forth movement through a balance or central position, so that the maximum displacement on one side of that position is equal to the maximum displacement on the other side. In the absence of any driving or damping force, natural frequency, also known as eigenfrequency, is the frequency at which a system tends to oscillate. The normal mode is called the motion pattern of a system that oscillates at its natural frequency (if all parts of the system move sinusoidally at the same frequency).
Formula used:
$\mathrm{T}=2 \pi \sqrt{\dfrac{\mathrm{I}}{\mathrm{mg} \mathrm{y}_{\mathrm{cm}}}}$
T = Time period
m = mass of the body
g= gravitational acceleration
y = Center of the mass (Y coordinate)
Complete Step-by-Step solution:
Objects can oscillate in all kinds of ways, but SHM and Simple Harmonic Motion are a really important form of oscillation. The object's acceleration is directly proportional to its displacement from its position in equilibrium.
The frequency (f) of an oscillation is measured in hertz (Hz) it is the number of oscillations per second. The time for one oscillation is called the period (T) it is measured in seconds.
$f=\dfrac{1}{T}$
$\mathrm{T}=2 \pi \sqrt{\dfrac{\mathrm{I}}{\mathrm{mg} \mathrm{y}_{\mathrm{cm}}}}$
$\mathrm{I}=\dfrac{\mathrm{MR}^{2}}{2}-\left(\dfrac{1}{2} \dfrac{\mathrm{M}}{4}\left(\dfrac{\mathrm{R}}{2}\right)^{2}+\dfrac{\mathrm{M}}{4}\left(\dfrac{\mathrm{R}}{2}\right)^{2}\right)$
$=\dfrac{13 \mathrm{MR}^{2}}{32} \mathrm{y}_{\mathrm{cm}}=\dfrac{\mathrm{R}}{6}$
$\therefore \text{T}=2\pi \sqrt{\dfrac{13}{4}\quad \dfrac{\text{R}}{\text{g}}}$
The time period of small oscillations of remaining portion about $O$ is $\pi \sqrt{\dfrac{13R}{g}}$
Hence, the correct option is (B).
Note:
In simple harmonic motion, repetitive back and forth movement through a balance or central position, so that the maximum displacement on one side of that position is equal to the maximum displacement on the other side. In the absence of any driving or damping force, natural frequency, also known as eigenfrequency, is the frequency at which a system tends to oscillate. The normal mode is called the motion pattern of a system that oscillates at its natural frequency (if all parts of the system move sinusoidally at the same frequency).
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