Energy in simple harmonic motion IB DP Physics Study Notes - 2025 Syllabus
Energy in simple harmonic motionmotion IB DP Physics Study Notes
Energy in simple harmonic motion IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
a qualitative approach to energy changes during one cycle of an oscillation
Standard level and higher level: 3 hours
Additional higher level: 4 hours
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Qualitatively describing the energy changes taking place during one cycle of an oscillation
- Consider the pendulum to the right which is placed in position and held there.
- Let the green rectangle represent the potential energy of the system.
- Let the red rectangle represent the kinetic energy of the system.
- Because there is no motion yet, there is no kinetic energy. But if we release it, the kinetic energy will grow as the potential energy diminishes.
- A continuous exchange between EK and EP occurs.
Consider the mass-spring system shown here. The mass is pulled to the right and held in place.
- Let the green rectangle represent the potential energy of the system.
- Let the red rectangle represent the kinetic energy of the system.
- A continuous exchange between EK and EP occurs.
- Note that the sum of EK and EP is constant.
- FYI ∙If friction and drag are both zero ET = CONST.
- If we plot both kinetic energy and potential energy vs. time for either system we would get the following graph
IB Physics Energy in simple harmonic motion Exam Style Worked Out Questions
Question
A mass oscillating in simple harmonic motion on the end of a spring has an amplitude \(x_0\) and a total energy \(E_{\mathrm{T}}\). The mass on the spring is doubled and made to oscillate with the same amplitude \(x_0\).
What is the total energy of the oscillating system after the change?
A. \(E_{\mathrm{T}}\)
B. \(\sqrt{2} E_{\mathrm{T}}\)
C. \(2 E_{\mathrm{T}}\)
D. \(4 E_{\mathrm{T}}\)
▶️Answer/Explanation
Ans:A
The total energy \(E_T\) of a simple harmonic oscillator is a constant and is proportional to the square of the amplitude (\(x_0\)) of the oscillation. Therefore, if the amplitude remains the same and only the mass on the spring is doubled, the total energy will also remain the same. This is because the total energy of the system is not dependent on the mass of the object undergoing simple harmonic motion.
Question
An object at the end of a spring oscillates vertically with simple harmonic motion (shm). The graph shows the variation with time t of the displacement x of the object.
What is the velocity of the object?
A -\(\frac{2\pi A}{T}sin(\frac{\pi t}{t})\)
B \(\frac{2\pi A}{T}sin(\frac{\pi t}{t})\)
C -\(\frac{2\pi A}{T}cos(\frac{\pi t}{t})\)
D \(\frac{2\pi A}{T}cos(\frac{\pi t}{t})\)
Answer/Explanation
Answer – A
Displacement \( =2A\sin \left ( \omega t-\frac{\pi }{2} \right ) \)
\( x=2A\cos \omega t \)
On differentiation
\( v=-2A\omega \sin \left ( \omega t \right ) \)
\( 2T=\frac{2\pi }{\omega } \)
\( \omega =\frac{\pi }{T} \)
\( v=-\frac{2A\pi }{T}\sin \left ( \frac{\pi }{T} t\right ) \)