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HPCET Oscillations — practice questions

Explore 75 free Oscillations questions in HPCET Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Oscillations questions and exercises with answers

Showing questions 1–30 of 75.

The time period of a simple pendulum of length $L$ in a gravitational field $g$ is:Read the answer and worked solution →A particle in SHM has amplitude $A$ and angular frequency $\omega$. Its maximum speed is:Read the answer and worked solution →The time period of a spring-mass system with mass $m$ and spring constant $k$ is:Read the answer and worked solution →The total mechanical energy of a particle of mass $m$ in SHM with amplitude $A$ and angular frequency $\omega$ is:Read the answer and worked solution →The time period of a simple pendulum of length $L$ is:Read the answer and worked solution →Time period of a spring-mass system (spring constant $k$, mass $m$) is:Read the answer and worked solution →At the **mean (equilibrium) position** of SHM, which is true?Read the answer and worked solution →Total mechanical energy of a particle in SHM (mass $m$, amplitude $A$, angular frequency $\omega$) is:Read the answer and worked solution →If displacement of a particle in SHM is $x = A\sin(\omega t)$, velocity at $x = A/2$ is:Read the answer and worked solution →In SHM, when the displacement is half the amplitude (x = A/2), the ratio of KE to PE is:Read the answer and worked solution →A simple pendulum of period $T$ on Earth is taken to the Moon where $g_{moon} = g/6$. Its new period is:Read the answer and worked solution →The phase difference between displacement $x$ and velocity $v$ in SHM is:Read the answer and worked solution →A spring of force constant $k$ is cut into two equal halves. The force constant of each half is:Read the answer and worked solution →Two springs of constants $k_1$ and $k_2$ connected in **series** have effective spring constant:Read the answer and worked solution →If $x_1 = A\sin(\omega t)$ and $x_2 = A\sin(\omega t + \pi/3)$, the resultant amplitude of $x_1 + x_2$ is:Read the answer and worked solution →Maximum acceleration in SHM ($A = $ amplitude, $\omega = $ angular frequency) is:Read the answer and worked solution →In a damped oscillator $\ddot x + 2\gamma\dot x + \omega_0^2 x = 0$, the **amplitude decays** as:Read the answer and worked solution →A particle in SHM has displacement $x = a\cos(\omega t) + b\sin(\omega t)$. The amplitude of motion is:Read the answer and worked solution →A particle performs SHM along x-axis with amplitude $A$ and period $T$. The time taken from $x = 0$ to $x = A/2$ is:Read the answer and worked solution →A particle in SHM has a maximum speed of 30 cm/s and a maximum acceleration of 90 cm/s². The period of oscillation is:Read the answer and worked solution →An SHM is given by $x = 5\sin(4\pi t + \pi/4)$ cm. The displacement at $t = 0.25$ s is:Read the answer and worked solution →A particle moving back and forth over the same path is undergoing:Read the answer and worked solution →In SHM, the time for one complete oscillation is called:Read the answer and worked solution →The general equation for SHM is:Read the answer and worked solution →In SHM, displacement and acceleration are:Read the answer and worked solution →The angular frequency ω of a spring-mass system with mass m and spring constant k is:Read the answer and worked solution →Time period of a simple pendulum of length L (small oscillations):Read the answer and worked solution →A spring-mass system has mass 4 kg and spring constant 100 N/m. Frequency of SHM:Read the answer and worked solution →For SHM x = 5 sin(2πt), find amplitude and time period:Read the answer and worked solution →Maximum velocity of a particle in SHM with amplitude A and angular frequency ω:Read the answer and worked solution →

Worked practice questions

The time period of a simple pendulum of length $L$ in a gravitational field $g$ is:

  1. $T = 2\pi \sqrt{\dfrac{L}{g}}$
  2. $T = 2\pi \sqrt{\dfrac{g}{L}}$
  3. $T = 2\pi L g$
  4. $T = \sqrt{\dfrac{L}{g}}$
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Correct answer: A. $T = 2\pi \sqrt{\dfrac{L}{g}}$

For small oscillations the pendulum's equation of motion reduces to SHM with $\omega = \sqrt{g/L}$. So period $T = 2\pi/\omega = 2\pi\sqrt{L/g}$. Independent of the mass of the bob (a key contrast with the spring-mass system, where $T \propto \sqrt{m}$). Note: only valid for small-amplitude swings. At larger amplitudes the small-angle approximation fails and the period depends slightly on amplitude.

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A particle in SHM has amplitude $A$ and angular frequency $\omega$. Its maximum speed is:

  1. $A\omega$
  2. $A\omega^2$
  3. $A/\omega$
  4. $\omega/A$
Show answer and solution

Correct answer: A. $A\omega$

Position $x(t) = A \sin(\omega t + \phi)$. Velocity is its time derivative: $v(t) = A\omega \cos(\omega t + \phi)$. Maximum value is $v_{\max} = A\omega$, achieved when the particle passes through the mean position ($x = 0$). Maximum acceleration is at the extremes: $a_{\max} = A\omega^2$ (option B is the acceleration max, the common trap).

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The time period of a spring-mass system with mass $m$ and spring constant $k$ is:

  1. $T = 2\pi \sqrt{\dfrac{k}{m}}$
  2. $T = 2\pi \sqrt{\dfrac{m}{k}}$
  3. $T = 2\pi k m$
  4. $T = 2\pi (k + m)$
Show answer and solution

Correct answer: B. $T = 2\pi \sqrt{\dfrac{m}{k}}$

Spring-mass system: Newton's second law gives $m\ddot x = -kx$, hence $\omega = \sqrt{k/m}$. Period: $T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{m/k}$. A stiffer spring (larger $k$) gives a shorter period. A heavier mass gives a longer period. Option A inverts the ratio, the classic trap.

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