Top 50 Most Repeated OSCILLATIONS PYQs | JEE MAINS
A curated collection of the most important questions from OSCILLATIONS, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from OSCILLATIONS, fully solved with step-by-step concepts to prepare for JEE MAINS.
For SHM, $a = -\omega^{2}x$. Converting $x$ to metres ($0.03\,\text{m}$) gives $a = - (10)^{2} \times 0.03 = -3\,\text{m s}^{-2}$. The negative sign i...
Read Full Step-by-Step Solution →$T \propto \sqrt{L}$, so 4× length → 2× period....
Read Full Step-by-Step Solution →At resonance in a forced damped oscillator, the driving frequency matches the system's resonant frequency (close to natural frequency). The phase diff...
Read Full Step-by-Step Solution →For a mass‑spring system, $T = 2\pi\sqrt{m/k}$. Rearranging gives $k = (4\pi^{2} m)/T^{2}$. Substituting $m=0.5\,\text{kg}$ and $T=0.5\,\text{s}$ yiel...
Read Full Step-by-Step Solution →First compute $\omega = 1/\sqrt{LC}=1/\sqrt{2\times10^{-3}\times0.5\times10^{-6}} = 3.16\times10^{4}\,\text{rad/s}$. Then $f = \omega/(2\pi) \approx 5...
Read Full Step-by-Step Solution →At resonance, driving frequency equals natural frequency for light damping....
Read Full Step-by-Step Solution →LC oscillations have $\omega = 1/\sqrt{LC}$, analogous to SHM's angular frequency....
Read Full Step-by-Step Solution →At $t=0$, $x = 5\sin(\pi/4) = 5 \times 0.707 \approx 3.54$....
Read Full Step-by-Step Solution →Effective $g$ increases to $g+a$. Since $T \propto 1/\sqrt{g}$, period decreases....
Read Full Step-by-Step Solution →First compute the spring constant $k = m\omega^{2} = 0.20\times10^{2}=20\,\text{N m}^{-1}$. Convert the amplitude to metres ($5\,\text{cm}=0.05\,\text...
Read Full Step-by-Step Solution →Using $T = 2\pi\sqrt{m/k}$, $T \approx 1.26\,\text{s}$....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Oscillations....
Read Full Step-by-Step Solution →For SHM, $a = -\omega^{2}x$. Converting $x$ to metres ($0.03\,\text{m}$) gives $a = - (10)^{2} \times 0.03 = -3\,\text{m s}^{-2}$. The negative sign i...
Read Full Step-by-Step Solution →$T \propto \sqrt{L}$, so 4× length → 2× period....
Read Full Step-by-Step Solution →At resonance in a forced damped oscillator, the driving frequency matches the system's resonant frequency (close to natural frequency). The phase diff...
Read Full Step-by-Step Solution →For a mass‑spring system, $T = 2\pi\sqrt{m/k}$. Rearranging gives $k = (4\pi^{2} m)/T^{2}$. Substituting $m=0.5\,\text{kg}$ and $T=0.5\,\text{s}$ yiel...
Read Full Step-by-Step Solution →First compute $\omega = 1/\sqrt{LC}=1/\sqrt{2\times10^{-3}\times0.5\times10^{-6}} = 3.16\times10^{4}\,\text{rad/s}$. Then $f = \omega/(2\pi) \approx 5...
Read Full Step-by-Step Solution →At resonance, driving frequency equals natural frequency for light damping....
Read Full Step-by-Step Solution →LC oscillations have $\omega = 1/\sqrt{LC}$, analogous to SHM's angular frequency....
Read Full Step-by-Step Solution →At $t=0$, $x = 5\sin(\pi/4) = 5 \times 0.707 \approx 3.54$....
Read Full Step-by-Step Solution →Effective $g$ increases to $g+a$. Since $T \propto 1/\sqrt{g}$, period decreases....
Read Full Step-by-Step Solution →First compute the spring constant $k = m\omega^{2} = 0.20\times10^{2}=20\,\text{N m}^{-1}$. Convert the amplitude to metres ($5\,\text{cm}=0.05\,\text...
Read Full Step-by-Step Solution →Using $T = 2\pi\sqrt{m/k}$, $T \approx 1.26\,\text{s}$....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Oscillations....
Read Full Step-by-Step Solution →For SHM, $a = -\omega^{2}x$. Converting $x$ to metres ($0.03\,\text{m}$) gives $a = - (10)^{2} \times 0.03 = -3\,\text{m s}^{-2}$. The negative sign i...
Read Full Step-by-Step Solution →$T \propto \sqrt{L}$, so 4× length → 2× period....
Read Full Step-by-Step Solution →At resonance in a forced damped oscillator, the driving frequency matches the system's resonant frequency (close to natural frequency). The phase diff...
Read Full Step-by-Step Solution →For a mass‑spring system, $T = 2\pi\sqrt{m/k}$. Rearranging gives $k = (4\pi^{2} m)/T^{2}$. Substituting $m=0.5\,\text{kg}$ and $T=0.5\,\text{s}$ yiel...
Read Full Step-by-Step Solution →First compute $\omega = 1/\sqrt{LC}=1/\sqrt{2\times10^{-3}\times0.5\times10^{-6}} = 3.16\times10^{4}\,\text{rad/s}$. Then $f = \omega/(2\pi) \approx 5...
Read Full Step-by-Step Solution →At resonance, driving frequency equals natural frequency for light damping....
Read Full Step-by-Step Solution →LC oscillations have $\omega = 1/\sqrt{LC}$, analogous to SHM's angular frequency....
Read Full Step-by-Step Solution →At $t=0$, $x = 5\sin(\pi/4) = 5 \times 0.707 \approx 3.54$....
Read Full Step-by-Step Solution →Effective $g$ increases to $g+a$. Since $T \propto 1/\sqrt{g}$, period decreases....
Read Full Step-by-Step Solution →First compute the spring constant $k = m\omega^{2} = 0.20\times10^{2}=20\,\text{N m}^{-1}$. Convert the amplitude to metres ($5\,\text{cm}=0.05\,\text...
Read Full Step-by-Step Solution →Using $T = 2\pi\sqrt{m/k}$, $T \approx 1.26\,\text{s}$....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Oscillations....
Read Full Step-by-Step Solution →For SHM, $a = -\omega^{2}x$. Converting $x$ to metres ($0.03\,\text{m}$) gives $a = - (10)^{2} \times 0.03 = -3\,\text{m s}^{-2}$. The negative sign i...
Read Full Step-by-Step Solution →$T \propto \sqrt{L}$, so 4× length → 2× period....
Read Full Step-by-Step Solution →At resonance in a forced damped oscillator, the driving frequency matches the system's resonant frequency (close to natural frequency). The phase diff...
Read Full Step-by-Step Solution →For a mass‑spring system, $T = 2\pi\sqrt{m/k}$. Rearranging gives $k = (4\pi^{2} m)/T^{2}$. Substituting $m=0.5\,\text{kg}$ and $T=0.5\,\text{s}$ yiel...
Read Full Step-by-Step Solution →First compute $\omega = 1/\sqrt{LC}=1/\sqrt{2\times10^{-3}\times0.5\times10^{-6}} = 3.16\times10^{4}\,\text{rad/s}$. Then $f = \omega/(2\pi) \approx 5...
Read Full Step-by-Step Solution →At resonance, driving frequency equals natural frequency for light damping....
Read Full Step-by-Step Solution →LC oscillations have $\omega = 1/\sqrt{LC}$, analogous to SHM's angular frequency....
Read Full Step-by-Step Solution →At $t=0$, $x = 5\sin(\pi/4) = 5 \times 0.707 \approx 3.54$....
Read Full Step-by-Step Solution →Effective $g$ increases to $g+a$. Since $T \propto 1/\sqrt{g}$, period decreases....
Read Full Step-by-Step Solution →First compute the spring constant $k = m\omega^{2} = 0.20\times10^{2}=20\,\text{N m}^{-1}$. Convert the amplitude to metres ($5\,\text{cm}=0.05\,\text...
Read Full Step-by-Step Solution →Using $T = 2\pi\sqrt{m/k}$, $T \approx 1.26\,\text{s}$....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Oscillations....
Read Full Step-by-Step Solution →For SHM, $a = -\omega^{2}x$. Converting $x$ to metres ($0.03\,\text{m}$) gives $a = - (10)^{2} \times 0.03 = -3\,\text{m s}^{-2}$. The negative sign i...
Read Full Step-by-Step Solution →$T \propto \sqrt{L}$, so 4× length → 2× period....
Read Full Step-by-Step Solution →