
A mass m oscillates with simple harmonic motion with frequency $f = \dfrac{\omega }{{2\pi }}$ and amplitude A on a spring with constant K, therefore
A. The total energy of the system is $\dfrac{1}{2}K{A^2}$
B. The frequency is $\dfrac{1}{{2\pi }}\sqrt {\dfrac{K}{m}} $
C. The maximum velocity occurs, when x=0
D. All the above are correct
Answer
164.7k+ views
Hint:The question is based on the waves and oscillation section of physics. We need to apply the concept of simple harmonic motion to find the correct options.
Complete step by step solution:
Simple harmonic motion (SHM) is defined as an oscillatory motion where the particle's acceleration at any position is inversely proportional to its displacement from the mean position. Conditions of simple harmonic motion are
$F \propto - x$
$\Rightarrow a \propto - x$
Where F = restoring force, a = acceleration and x = displacement of the particle from the mean position.
The total energy of the particle undergoing simple harmonic motion is given below.
$E = K + V$
$\Rightarrow E = \left[ {\dfrac{1}{2}m{\omega ^2}({A^2} - {x^2})} \right] + \left[ {\dfrac{1}{2}m{\omega ^2}{x^2}} \right]$
$\Rightarrow E = \dfrac{1}{2}m{\omega ^2}{A^2}$
Where $E$= total energy, $K$= kinetic energy, $V$= potential energy, $m$ = mass, $x$= displacement from mean position, $A$= amplitude, $\omega $ = angular frequency.
The equation for the frequency of particles undergoing simple harmonic motion is given below.
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{K}{m}} $
Where $f$= frequency, $K$= force constant, $m$=mass.
The equation for the velocity of a particle undergoing simple harmonic motion is given below.
$v = \omega \sqrt {{A^2} - {x^2}} $
$\Rightarrow {v_{\max }} = \omega \sqrt {{A^2}} $ with $x = 0$
Where $v$= velocity, $A$= amplitude, $\omega $ = angular frequency and $x$= displacement from mean position.
So, all three of the statements are correct.
Hence, the correct option is option D.
Note: All the SHM are oscillatory and periodic, but not all oscillatory motions are SHM. Understanding the fundamentals of the simple harmonic motion is crucial to understanding the properties of alternating currents, light waves, and sound waves. Simple Harmonic Motion can be classified into two types: linear SHM and angular SHM.
Complete step by step solution:
Simple harmonic motion (SHM) is defined as an oscillatory motion where the particle's acceleration at any position is inversely proportional to its displacement from the mean position. Conditions of simple harmonic motion are
$F \propto - x$
$\Rightarrow a \propto - x$
Where F = restoring force, a = acceleration and x = displacement of the particle from the mean position.
The total energy of the particle undergoing simple harmonic motion is given below.
$E = K + V$
$\Rightarrow E = \left[ {\dfrac{1}{2}m{\omega ^2}({A^2} - {x^2})} \right] + \left[ {\dfrac{1}{2}m{\omega ^2}{x^2}} \right]$
$\Rightarrow E = \dfrac{1}{2}m{\omega ^2}{A^2}$
Where $E$= total energy, $K$= kinetic energy, $V$= potential energy, $m$ = mass, $x$= displacement from mean position, $A$= amplitude, $\omega $ = angular frequency.
The equation for the frequency of particles undergoing simple harmonic motion is given below.
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{K}{m}} $
Where $f$= frequency, $K$= force constant, $m$=mass.
The equation for the velocity of a particle undergoing simple harmonic motion is given below.
$v = \omega \sqrt {{A^2} - {x^2}} $
$\Rightarrow {v_{\max }} = \omega \sqrt {{A^2}} $ with $x = 0$
Where $v$= velocity, $A$= amplitude, $\omega $ = angular frequency and $x$= displacement from mean position.
So, all three of the statements are correct.
Hence, the correct option is option D.
Note: All the SHM are oscillatory and periodic, but not all oscillatory motions are SHM. Understanding the fundamentals of the simple harmonic motion is crucial to understanding the properties of alternating currents, light waves, and sound waves. Simple Harmonic Motion can be classified into two types: linear SHM and angular SHM.
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