SHM flash cards

1. 1.Define Simple Harmonic Motion (SHM).

   SHM is a special periodic (to-and-fro) oscillatory motion in which the restoring force (or torque) is directly proportional to the displacement from the mean position and always directed towards it: $F = -kx$.

   Hint: Restoring force $\propto$ displacement, opposite in direction.

2. 2.What is the defining differential equation of linear SHM?

   $\dfrac{d^2x}{dt^2} = -\omega^2 x$, i.e. acceleration $a = -\omega^2 x$. Any motion obeying this is SHM with angular frequency $\omega$.

   Hint: Second derivative $\propto$ negative of the variable.

3. 3.What are the necessary conditions for a body to execute SHM?

   (1) There must be a stable equilibrium (mean) position, (2) a restoring force directed towards it, and (3) this force must be linearly proportional to displacement, $F = -kx$.

   Hint: Stable equilibrium + linear restoring force.

4. 4.Write the general displacement equation of SHM and name each symbol.

   $x = A\sin(\omega t + \phi)$, where $A$ = amplitude, $\omega$ = angular frequency, $t$ = time, $\phi$ = initial phase (epoch), and $(\omega t+\phi)$ = phase.

   Hint: Sine (or cosine) with amplitude, angular frequency, phase.

5. 5.How are angular frequency $\omega$, time period $T$ and frequency $f$ related?

   $\omega = \dfrac{2\pi}{T} = 2\pi f$, so $T = \dfrac{2\pi}{\omega}$ and $f = \dfrac{1}{T} = \dfrac{\omega}{2\pi}$.

   Hint: $\omega = 2\pi f$.

6. 6.For $x = A\sin(\omega t+\phi)$, write the velocity as a function of time.

   $v = \dfrac{dx}{dt} = A\omega\cos(\omega t+\phi)$. Maximum speed $v_{max} = A\omega$ at the mean position.

   Hint: Differentiate displacement once.

7. 7.Express velocity in SHM as a function of displacement $x$.

   $v = \pm\,\omega\sqrt{A^2 - x^2}$. It is maximum ($\omega A$) at $x=0$ and zero at $x=\pm A$.

   Hint: Eliminate $t$ between $x$ and $v$.

8. 8.For $x = A\sin(\omega t+\phi)$, write the acceleration and its maximum value.

   $a = -A\omega^2\sin(\omega t+\phi) = -\omega^2 x$. Maximum magnitude $a_{max} = A\omega^2$ at the extreme positions.

   Hint: Differentiate velocity; $a=-\omega^2 x$.

9. 9.Where in SHM are velocity and acceleration maximum and minimum?

   Velocity is max at mean position ($x=0$) and zero at extremes. Acceleration is zero at mean position and max at extremes ($x=\pm A$).

   Hint: They peak at opposite ends of the path.

10. 10.What is the phase difference between displacement, velocity and acceleration in SHM?

    Velocity leads displacement by $\dfrac{\pi}{2}$; acceleration leads displacement by $\pi$ (out of phase). Acceleration leads velocity by $\dfrac{\pi}{2}$.

    Hint: Each derivative advances phase by $90^\circ$.

11. 11.Sketch/describe the acceleration–displacement graph for SHM.

    It is a straight line through the origin with negative slope: $a = -\omega^2 x$. The slope magnitude equals $\omega^2$.

    Hint: Straight line, slope $=-\omega^2$.

12. 12.Describe the velocity–displacement curve for SHM.

    $\dfrac{v^2}{\omega^2 A^2} + \dfrac{x^2}{A^2} = 1$ — an ellipse in the $v$–$x$ plane with semi-axes $A$ (along $x$) and $\omega A$ (along $v$).

    Hint: $v=\omega\sqrt{A^2-x^2}$ squared gives an ellipse.

13. 13.What is meant by the phase and epoch (initial phase) of an SHM?

    Phase $= (\omega t + \phi)$ specifies the state (position and direction) at time $t$. The epoch $\phi$ is the phase at $t=0$, fixing the starting configuration.

    Hint: Argument of the sine; its value at $t=0$.

14. 14.How is SHM related to uniform circular motion?

    SHM is the projection of uniform circular motion (radius $A$, angular speed $\omega$) onto a diameter. The foot of the perpendicular from the particle executes SHM with amplitude $A$ and angular frequency $\omega$.

    Hint: Shadow of a particle moving in a circle.

15. 15.Write the potential energy of a particle in SHM as a function of displacement.

    $U = \dfrac{1}{2}kx^2 = \dfrac{1}{2}m\omega^2 x^2$ (taking $U=0$ at mean position). It is maximum at extremes.

    Hint: $U\propto x^2$, parabola.

16. 16.Write the kinetic energy of a particle in SHM as a function of displacement.

    $K = \dfrac{1}{2}m v^2 = \dfrac{1}{2}m\omega^2 (A^2 - x^2)$. Maximum at mean position, zero at extremes.

    Hint: $K\propto (A^2-x^2)$.

17. 17.What is the total mechanical energy of a particle in SHM?

    $E = K + U = \dfrac{1}{2}m\omega^2 A^2 = \dfrac{1}{2}kA^2$, a constant independent of $x$. It depends on $A^2$ and $\omega^2$.

    Hint: Energy $\propto A^2$; constant everywhere.

18. 18.How does the total energy of SHM depend on amplitude and frequency?

    $E = \dfrac{1}{2}m\omega^2 A^2$, so $E \propto A^2$ and $E \propto \omega^2 \propto f^2$. Doubling amplitude quadruples energy.

    Hint: $E\propto A^2 f^2$.

19. 19.At what displacement are the kinetic and potential energies of an SHM equal?

    When $K = U$: $\dfrac12 m\omega^2(A^2-x^2) = \dfrac12 m\omega^2 x^2 \Rightarrow x = \pm\dfrac{A}{\sqrt2}$. Each equals $E/2$ there.

    Hint: Set $A^2 - x^2 = x^2$.

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