Answer
Verified
111.6k+ views
Hint: The diagram given to us is the phasor diagram for the simple harmonic motion of the oscillator. The radius of the circle gives us the amplitude of oscillations of the oscillator and the angular velocity for the circle gives us the frequency of oscillation of the oscillator. We need to find the mechanical energy of the oscillator in the two cases.
Formula Used:
$\implies$ \[K=\dfrac{1}{2}m{{v}^{2}}\]
$\implies$ \[P=m\times v\]
$\implies$ \[K=\dfrac{{{P}^{2}}}{2m}\]
Complete step by step answer:
In case of simple harmonic motion, at the mean position of oscillation, the total mechanical energy appears in the form of kinetic energy and at the extreme positions, the total mechanical energy appears in the form of potential energy. Hence, the total mechanical energy of the oscillator is equal to the kinetic energy in the mean position.
The mean position of the oscillator is shown by the y-axis in the given figure. Since the momentum of the two cases is marked along the y-axis, we can say that \[{{P}_{1}}=2{{P}_{2}}\] where \[P\] denotes momentum of the oscillator.
From the equations \[K=\dfrac{1}{2}m{{v}^{2}}\] and \[P=m\times v\] , we can say that \[K=\dfrac{{{P}^{2}}}{2m}\] where \[P\] denotes momentum and \[K\] denotes kinetic energy of the particle in oscillation
Now, substituting the values of momentum for the two cases, we have
\[{{E}_{1}}=\dfrac{{{(2{{P}_{1}})}^{2}}}{2m}\] and \[{{E}_{2}}=\dfrac{{{P}_{1}}^{2}}{2m}\] where \[{{E}_{1}}\] and \[{{E}_{2}}\] are the total mechanical energies of the oscillations
Taking the ratio of the two energies obtained, we get
\[\dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{4}{1}\Rightarrow {{E}_{1}}=4{{E}_{2}}\]
Hence option (C) is the correct answer.
Note: We can alternatively solve this question with the help of the relation of kinetic energy in simple harmonic motion and the amplitude of the oscillation, that is \[K=\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] where \[K\] is the kinetic energy, \[\omega \] is the frequency of oscillation and \[A\] is the amplitude of oscillation. If we approach the question using this method, we won’t even need the diagram. We can just use the statement that their amplitudes are in the ratio \[2:1\] .
Formula Used:
$\implies$ \[K=\dfrac{1}{2}m{{v}^{2}}\]
$\implies$ \[P=m\times v\]
$\implies$ \[K=\dfrac{{{P}^{2}}}{2m}\]
Complete step by step answer:
In case of simple harmonic motion, at the mean position of oscillation, the total mechanical energy appears in the form of kinetic energy and at the extreme positions, the total mechanical energy appears in the form of potential energy. Hence, the total mechanical energy of the oscillator is equal to the kinetic energy in the mean position.
The mean position of the oscillator is shown by the y-axis in the given figure. Since the momentum of the two cases is marked along the y-axis, we can say that \[{{P}_{1}}=2{{P}_{2}}\] where \[P\] denotes momentum of the oscillator.
From the equations \[K=\dfrac{1}{2}m{{v}^{2}}\] and \[P=m\times v\] , we can say that \[K=\dfrac{{{P}^{2}}}{2m}\] where \[P\] denotes momentum and \[K\] denotes kinetic energy of the particle in oscillation
Now, substituting the values of momentum for the two cases, we have
\[{{E}_{1}}=\dfrac{{{(2{{P}_{1}})}^{2}}}{2m}\] and \[{{E}_{2}}=\dfrac{{{P}_{1}}^{2}}{2m}\] where \[{{E}_{1}}\] and \[{{E}_{2}}\] are the total mechanical energies of the oscillations
Taking the ratio of the two energies obtained, we get
\[\dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{4}{1}\Rightarrow {{E}_{1}}=4{{E}_{2}}\]
Hence option (C) is the correct answer.
Note: We can alternatively solve this question with the help of the relation of kinetic energy in simple harmonic motion and the amplitude of the oscillation, that is \[K=\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] where \[K\] is the kinetic energy, \[\omega \] is the frequency of oscillation and \[A\] is the amplitude of oscillation. If we approach the question using this method, we won’t even need the diagram. We can just use the statement that their amplitudes are in the ratio \[2:1\] .
Latest Vedantu courses for you
Grade 11 Science PCM | CBSE | SCHOOL | English
CBSE (2024-25)
Academic year 2024-25
ENGLISH
Unlimited access till final school exam
School Full course for CBSE students
Physics
Chemistry
Maths
₹35,000 per year
EMI starts from ₹2,916.67 per month
Recently Updated Pages
Write an article on the need and importance of sports class 10 english JEE_Main
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
If x2 hx 21 0x2 3hx + 35 0h 0 has a common root then class 10 maths JEE_Main
The radius of a sector is 12 cm and the angle is 120circ class 10 maths JEE_Main
For what value of x function fleft x right x4 4x3 + class 10 maths JEE_Main
Other Pages
Radius of the largest circle which passes through -class-11-maths-JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
An aeroplane left 50 minutes later than its schedu-class-11-maths-JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main