An Angular Simple Harmonic Oscillator MCQ Quiz - Objective Question with Answer for An Angular Simple Harmonic Oscillator - Download Free PDF

$M^{\prime }=4m$

$M\to$ magnetic moment of $1^{st}$ magnet

$M^{\prime }\to$ magnetic moment of now magnet

$T=2\pi \sqrt{{\displaystyle \frac{I}{M{B}_H}}}$

$T\propto {\displaystyle \frac{1}{\sqrt{M}}}$

$\frac{T^{\prime }}{T}=\sqrt{{\displaystyle \frac{M}{M^{\prime }}}}\Rightarrow {\displaystyle \frac{T^{\prime }}{2}}=\sqrt{{\displaystyle \frac{M}{4M}}}$

$\frac{T^{\prime }}{2}=1/2$

T' = 1

Concept:

• An ideal simple pendulum consists of a heavy point mass body suspended by a weightless, inextensible, and perfectly flexed supportable string from rigid support about which it is free to oscillate.
• Mathematically, the time period of swing of a pendulum,
  • ​T = 2π√l\g

Calculation: 

• For a simple pendulum, the time period of the simple pendulum and acceleration depends on the length of the string due to gravity,
  • ​T = 2π√l\g
• When its length is increased four times, then the time period of the simple pendulum
  • ​​T'= 2π√l\g = 2π√4l\g = 2T

Thus, the time period of a simple pendulum becomes twice. if its length is increased four times.

CONCEPT

• Torsional Pendulum: A torsional pendulum consists of a disk (or some other object) suspended from a wire, which is then twisted and released, resulting in an oscillatory motion. The oscillatory motion is caused by a restoring torque which is proportional to the angular displacement.

EXPLANATION

• If the torsional constant of the wire is k and the body is rotated through an angle θ, then the torque produced is $\tau =-k\theta$
• If I be the moment of inertia of the body about the verticle axis then the angular acceleration is $\alpha =\frac{\tau }{I}=-\frac{k}{I}\theta =-{\omega }^2\theta$
• where, $\omega =\sqrt{\frac{k}{I}}$

$\Rightarrow Timeperiod\left(T\right)=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{k}}$

$\Rightarrow Frequency\left(\frac{1}{T}\right)=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{k}{I}}$

⇒ Option 4 is correct

Concept:

• An ideal simple pendulum consists of a heavy point mass body suspended by a weightless, inextensible, and perfectly flexed supportable string from rigid support about which it is free to oscillate.
• Mathematically, the time period of swing of a pendulum,
  • ​T = 2π√l\g

Calculation: 

• For a simple pendulum, the time period of the simple pendulum and acceleration depends on the length of the string due to gravity,
  • ​T = 2π√l\g
• When its length is increased four times, then the time period of the simple pendulum
  • ​​T'= 2π√l\g = 2π√4l\g = 2T

Thus, the time period of a simple pendulum becomes twice. if its length is increased four times.

CONCEPT

• Torsional Pendulum: A torsional pendulum consists of a disk (or some other object) suspended from a wire, which is then twisted and released, resulting in an oscillatory motion. The oscillatory motion is caused by a restoring torque which is proportional to the angular displacement.

EXPLANATION

• If the torsional constant of the wire is k and the body is rotated through an angle θ, then the torque produced is $\tau =-k\theta$
• If I be the moment of inertia of the body about the verticle axis then the angular acceleration is $\alpha =\frac{\tau }{I}=-\frac{k}{I}\theta =-{\omega }^2\theta$
• where, $\omega =\sqrt{\frac{k}{I}}$

$\Rightarrow Timeperiod\left(T\right)=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{k}}$

$\Rightarrow Frequency\left(\frac{1}{T}\right)=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{k}{I}}$

⇒ Option 4 is correct

Concept:

• An ideal simple pendulum consists of a heavy point mass body suspended by a weightless, inextensible, and perfectly flexed supportable string from rigid support about which it is free to oscillate.
• Mathematically, the time period of swing of a pendulum,
  • ​T = 2π√l\g

Calculation: 

• For a simple pendulum, the time period of the simple pendulum and acceleration depends on the length of the string due to gravity,
  • ​T = 2π√l\g
• When its length is increased four times, then the time period of the simple pendulum
  • ​​T'= 2π√l\g = 2π√4l\g = 2T

Thus, the time period of a simple pendulum becomes twice. if its length is increased four times.

CONCEPT

• Torsional Pendulum: A torsional pendulum consists of a disk (or some other object) suspended from a wire, which is then twisted and released, resulting in an oscillatory motion. The oscillatory motion is caused by a restoring torque which is proportional to the angular displacement.

EXPLANATION

• If the torsional constant of the wire is k and the body is rotated through an angle θ, then the torque produced is $\tau =-k\theta$
• If I be the moment of inertia of the body about the verticle axis then the angular acceleration is $\alpha =\frac{\tau }{I}=-\frac{k}{I}\theta =-{\omega }^2\theta$
• where, $\omega =\sqrt{\frac{k}{I}}$

$\Rightarrow Timeperiod\left(T\right)=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{k}}$

$\Rightarrow Frequency\left(\frac{1}{T}\right)=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{k}{I}}$

⇒ Option 4 is correct

$M^{\prime }=4m$

$M\to$ magnetic moment of $1^{st}$ magnet

$M^{\prime }\to$ magnetic moment of now magnet

$T=2\pi \sqrt{{\displaystyle \frac{I}{M{B}_H}}}$

$T\propto {\displaystyle \frac{1}{\sqrt{M}}}$

$\frac{T^{\prime }}{T}=\sqrt{{\displaystyle \frac{M}{M^{\prime }}}}\Rightarrow {\displaystyle \frac{T^{\prime }}{2}}=\sqrt{{\displaystyle \frac{M}{4M}}}$

$\frac{T^{\prime }}{2}=1/2$

T' = 1