A particle is executing a simple harmonic motion. Its maximum acceleration is α and…
Physics · JEE Main · NTA Exams — Oscillations and Waves
A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then, its time period of vibration will be;-
- \(\frac{2 \pi \beta}{\pi}\)
- \(\frac{\beta^{2}}{a^{2}}\)
- \(\frac{\pi}{\beta}\)
- \(\frac{\beta^{2}}{x}\)
Answer
(A) 2
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- A cylindrical tube, open at both ends, has a fundamental frequency f in air. The tube is dipped vertically in…
- Two sources A and B are sounding notes of frequency 680 Hz. A listener moves from A to B with a constant…
- The expression for displacement of an object in SHM is x = A cos (ωt). The potential energy at t= T 2 is…
- Two particles are executing simple harmonic motion of the same amplitude A and frequency along the x–axis…
- An increase in intensity level of 1 dB implies an increase in Intensity of (given antilog10 0.1 = 1.2589)
- A tuning fork vibrating with a sonometer having 20 cm wire produces 5 beats/s. The beat frequency does not…
- The distance travelled by the wave pattern in the time required for one full oscillation by any constituent…
- A standing wave is formed on a string fixed at both the ends. The individual waves i.e., incident wave and…
More Oscillations and Waves questions · Browse all practice questions