Oscillations Previous Year Questions — NEET Physics

3 past-year Oscillations questions from NEET (Physics).

Q25 (2024)

For a simple pendulum, having time period T, the variation of kinetic energy (K.E.) with time (t) is represented by:
  1. Option (1)
  2. Option (2)
  3. Option (3)
  4. Option (4)
### Related Formula $$K = \frac{1}{2}m v^2 = \frac{1}{2}mA^{2}\omega^2 \cos^2(\omega t + \phi)$$ ### Core Logic Kinetic energy depends on $\cos^2(\omega t + \phi)$. $K.E.$ is non-negative and oscillates with frequency $2f$ (time period $T/2$). Curve in option (3) correctly represents $\cos^2$ variation. ### Pattern Recognition Kinetic energy in SHM varies sinusoidally with period T/2 and remains positive. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

Q32 (2024)

Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum L, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as : (Take $\pi^2 = 9.8$ , and $\mathrm{g} = 9.8\mathrm{m / s}^2$ )
  1. $2\mathrm{m}$
  2. $0.75\mathrm{m}$
  3. $1.5\mathrm{m}$
  4. $1\mathrm{m}$
### Related Formula $$T = 2\pi \sqrt{\frac{\ell}{g}} \Rightarrow \ell = \frac{g T^2}{4\pi^2}$$ ### Core Logic Time period $T = \frac{60}{30} = 2\text{ s}$. $$\ell = \frac{9.8 \times 2^2}{4 \times 9.8} = 1\mathrm{m}$$ ### Pattern Recognition A simple pendulum with T = 2s has length 1m when g = pi^2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

Q36 (2024)

The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately: (Consider mass of the bob = 20 g)
  1. 1.41 m/s
  2. 14.1 m/s
  3. 0.2 m/s
  4. 2.0 m/s
### Related Formula $$E_{\text{total}} = \frac{1}{2} m v_{\text{max}}^2$$ ### Core Logic At equilibrium position, potential energy is zero and total energy equals kinetic energy. $$\frac{1}{2} m v^2 = 0.02 \Rightarrow \frac{1}{2} \times 20 \times 10^{-3} \times v^2 = 0.02$$ $$10^{-2} v^2 = 2 \times 10^{-2} \Rightarrow v^2 = 2 \Rightarrow v = \sqrt{2} \approx 1.41 \text{ m/s}$$ ### Pattern Recognition Total energy equals maximum kinetic energy at equilibrium. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations
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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

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