A bar magnet is oscillating in the earth's magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled?
A. Motion remains SH with time period = $4T$
B. Motion remains SH and period remains nearly constant
C. Motion remains SH with time period = $\dfrac{\text{T}}{\text{2}}$
D. Motion remains SH with time period = $2T$
Answer
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Hint:Simple harmonic motion (abbreviated SHM) is a form of periodic motion in mechanics and physics in which the restoring force on a moving item is directly proportional to the size of the object's displacement and operates towards the object's equilibrium position. It causes an oscillation that, if not interrupted by friction or other energy dissipation, can last eternally. We use this concept here.
Formula used:
\[T=2\pi \sqrt{\dfrac{m}{k}}\]
Where, $T$ = time period and $M$ = mass of the system
Complete step by step answer:
Simple harmonic motion may be used to simulate a number of movements, but it is best shown by the oscillation of a mass on a spring when it is subjected to Hooke's law's linear elastic restoring force. The motion has a single resonance frequency and is sinusoidal in time.
Other phenomena, such as the motion of a simple pendulum, may be described using basic harmonic motion, however the net force on the item at the end of the pendulum must be proportional to the displacement for it to be an accurate model. Molecular vibration can also be modelled using simple harmonic motion.
Now in \[T=2\pi \sqrt{\dfrac{m}{k}}\] we can deduce that $\text{T }\!\!\alpha\!\!\text{ }\sqrt{m}$
So if mass is quadrupled
$\dfrac{{{T}_{1}}}{\sqrt{{{m}_{1}}}}=\dfrac{{{T}_{2}}}{\sqrt{{{m}_{2}}}}$
Given${{m}_{2}}=4{{m}_{1}}$
So
$\dfrac{{{T}_{1}}}{\sqrt{{{m}_{1}}}}=\dfrac{{{T}_{2}}}{\sqrt{4{{m}_{1}}}}$
Hence
$\therefore {{\text{T}}_{\text{2}}}\text{ = 2 }{{\text{T}}_{\text{1}}}$
So option D is correct.
Note:Always try to remember the formula of time period, also note that ${{m}_{2}}=4{{m}_{1}}$ Is the correct substitution and not $4{{m}_{2}}={{m}_{1}}$.Simple harmonic motion is the motion of a particle travelling down a straight line with an acceleration that is proportional to the distance from a fixed point on the line and whose magnitude is proportional to the distance from the fixed point.
Formula used:
\[T=2\pi \sqrt{\dfrac{m}{k}}\]
Where, $T$ = time period and $M$ = mass of the system
Complete step by step answer:
Simple harmonic motion may be used to simulate a number of movements, but it is best shown by the oscillation of a mass on a spring when it is subjected to Hooke's law's linear elastic restoring force. The motion has a single resonance frequency and is sinusoidal in time.
Other phenomena, such as the motion of a simple pendulum, may be described using basic harmonic motion, however the net force on the item at the end of the pendulum must be proportional to the displacement for it to be an accurate model. Molecular vibration can also be modelled using simple harmonic motion.
Now in \[T=2\pi \sqrt{\dfrac{m}{k}}\] we can deduce that $\text{T }\!\!\alpha\!\!\text{ }\sqrt{m}$
So if mass is quadrupled
$\dfrac{{{T}_{1}}}{\sqrt{{{m}_{1}}}}=\dfrac{{{T}_{2}}}{\sqrt{{{m}_{2}}}}$
Given${{m}_{2}}=4{{m}_{1}}$
So
$\dfrac{{{T}_{1}}}{\sqrt{{{m}_{1}}}}=\dfrac{{{T}_{2}}}{\sqrt{4{{m}_{1}}}}$
Hence
$\therefore {{\text{T}}_{\text{2}}}\text{ = 2 }{{\text{T}}_{\text{1}}}$
So option D is correct.
Note:Always try to remember the formula of time period, also note that ${{m}_{2}}=4{{m}_{1}}$ Is the correct substitution and not $4{{m}_{2}}={{m}_{1}}$.Simple harmonic motion is the motion of a particle travelling down a straight line with an acceleration that is proportional to the distance from a fixed point on the line and whose magnitude is proportional to the distance from the fixed point.
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