
A simple harmonic oscillator is of mass $0.100\,\,Kg$. It is oscillating with a frequency of $\dfrac{5}{\pi }\,\,Hz$. If its amplitude of vibration is $5\,\,cm$, the force acting on the particle at its extreme position is
(A) $2\,\,N$
(B) $1.5\,\,N$
(C) $1\,N$
(D) $0.5\,\,N$
Answer
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Hint:The above problem can be solved using the formula for the simple harmonic motion of an oscillator of a particular particle, which incorporates the frequency of vibration of that particle, mass of the oscillator and the amplitude of the vibration produced.
Formulae Used:
The force acting on the particle due to simple harmonic motion is;
$F = m{\omega ^2}A$
Where, $F$ denotes the force acting on the particle, $m$ denotes the mass of the oscillator, $\omega $ denotes the angular frequency, $A$ denotes the amplitude of vibration.
Complete step-by-step solution:
The data given in the problem is;
Mass of the oscillator is, $m = 0.100\,\,Kg$.
Oscillating frequency, $f = \dfrac{5}{\pi }\,\,Hz$.
Amplitude of vibration,
$A = 5\,\,cm$.
$A = 0.05\,\,m$
The force acting on the particle due to simple harmonic motion is;
$F = m{\omega ^2}A\,\,..........\left( 1 \right)$
We know that $\omega = 2\pi f$;
By substituting the value of frequency;
$\omega = 2\pi \times \dfrac{5}{\pi }$
By simplifying the above equation, we get;
$\omega = 10\,\,rad\,{s^{ - 1}}$
Substitute the value of $\omega $ in the equation (1);
$F = 0.1\,\,Kg \times {10^2}\,\,rad\,{s^{ - 1}} \times 0.05\,\,m$
On simplifying the above equation, we get;
$F = 0.1 \times 100 \times 0.05$
$F = 0.5\,\,N$
Therefore, the force acting on the particle due to simple harmonic motion in the extreme position is, $F = 0.5\,\,N$.
Hence the option (D) $F = 0.5\,\,N$ is the correct answer.
Additional information:
In the inclusion to linear motion and rotational motion there is a different kind of motion. This motion consists of the to and fro motion of swinging or vibrating. When a material oscillates, it travels back and forth with respect to time. It is very useful to trace out the location of an oscillating particle with time. This is known as the Simple Harmonic Motion (SHM).
Note:- The change in the mass of the oscillator in a Simple Harmonic motion the oscillation of frequency varies a great deal, therefore, the amplitude of the vibration also varies. While attending these problems do not fail to calculate the values that are in power as well.
Formulae Used:
The force acting on the particle due to simple harmonic motion is;
$F = m{\omega ^2}A$
Where, $F$ denotes the force acting on the particle, $m$ denotes the mass of the oscillator, $\omega $ denotes the angular frequency, $A$ denotes the amplitude of vibration.
Complete step-by-step solution:
The data given in the problem is;
Mass of the oscillator is, $m = 0.100\,\,Kg$.
Oscillating frequency, $f = \dfrac{5}{\pi }\,\,Hz$.
Amplitude of vibration,
$A = 5\,\,cm$.
$A = 0.05\,\,m$
The force acting on the particle due to simple harmonic motion is;
$F = m{\omega ^2}A\,\,..........\left( 1 \right)$
We know that $\omega = 2\pi f$;
By substituting the value of frequency;
$\omega = 2\pi \times \dfrac{5}{\pi }$
By simplifying the above equation, we get;
$\omega = 10\,\,rad\,{s^{ - 1}}$
Substitute the value of $\omega $ in the equation (1);
$F = 0.1\,\,Kg \times {10^2}\,\,rad\,{s^{ - 1}} \times 0.05\,\,m$
On simplifying the above equation, we get;
$F = 0.1 \times 100 \times 0.05$
$F = 0.5\,\,N$
Therefore, the force acting on the particle due to simple harmonic motion in the extreme position is, $F = 0.5\,\,N$.
Hence the option (D) $F = 0.5\,\,N$ is the correct answer.
Additional information:
In the inclusion to linear motion and rotational motion there is a different kind of motion. This motion consists of the to and fro motion of swinging or vibrating. When a material oscillates, it travels back and forth with respect to time. It is very useful to trace out the location of an oscillating particle with time. This is known as the Simple Harmonic Motion (SHM).
Note:- The change in the mass of the oscillator in a Simple Harmonic motion the oscillation of frequency varies a great deal, therefore, the amplitude of the vibration also varies. While attending these problems do not fail to calculate the values that are in power as well.
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