Simple Harmonic Motion Formulas

1. Simple harmonic motion

A oscillatory motion in which the restoring force is proportional to displacement and directed opposite to it.

2. Restoring force

F = -kx

3. Restoring couple

τ = -Cθ

4. Equation of motion

Linear S.H.M.
\(\frac{d^{2} x}{d t^{2}}\) + ╧ë²x = 0, where ╧ë² = \(\frac{k}{m}\)

Angular S.H.M.
\(\frac{d^{2} \theta}{d t^{2}}\) + ╧ë²╬╕ = 0, where ╧ë² = \(\frac{C}{I}\)

5. Displacement in S.H.M.

x = a sin (ωt + Φ)
where a = amplitude, ω = angular frequency = 2πn = \(\frac{2 \pi}{\mathrm{T}}\)
(ωt + Φ) = phase at time t, Φ = initial phase angle or phase constant.

6. Velocity in S.H.M.

v = \(\frac{d x}{d t}\) aω cos(ωt + Φ)
or v = ╧ë(a² – x²)^1/2
v_max = aω, when x = 0
& v_min = 0, when x = a

7. Acceleration in S.H.M.

f = \(\frac{d v}{d t}=\frac{d^{2} x}{d t^{2}}\) = -╧ë² a sin (╧ët + ╬ª)
or f = -╧ë²x
f_max = ╧ë²a, at extreme position
& f_min = 0, when x = 0

8. Period and frequency

Linear S.H.M.
T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{k}}\)
& frequency n = \(\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\)
Angular S.H.M.
T = \(2 \pi \sqrt{\frac{I}{C}}\), n = \(\frac{1}{2 \pi} \sqrt{\frac{C}{I}}\)

9. Kinetic energy in S.H.M.

KE = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) m╧ë²a²cos² (╧ët + ╬ª)
= \(\frac{1}{2}\) ka² cos² (╧ët + ╬ª)
or KE = \(\frac{1}{2}\)m╧ë²(a² – x²) = \(\frac{1}{2}\) k(a² – x²)
(KE)_max = \(\frac{1}{2}\) ka² = \(\frac{1}{2}\)m╧ë²a² when x = 0.
(KE)_min = 0. when x = ┬▒ a.
Mean KE with respect to displacement = \(\frac{1}{3}\) ka²
Mean KE with respect to time = \(\frac{1}{4}\) ka².

10. Potential energy in S.H.M.

PE = U(x) = \(\frac{1}{2}\) kx² = \(\frac{1}{2}\) m╧ë²x²
= \(\frac{1}{2}\) ka² sin² (╧ët + ╬ª)
(PE)_max = \(\frac{1}{2}\) ka² = \(\frac{1}{2}\) m╧ë²a²
= 2╧Dzmn²a², when x = ┬▒ a.
(PE)_min = 0, when x = 0.
Mean PE with respect to displacement = \(\frac{1}{6}\) ka²
Mean PE with respect to time = \(\frac{1}{4}\) ka².

11. Total energy in S.H.M

Total Energy E = KE + PE
\(\frac{1}{2}\) ka² = \(\frac{1}{2}\) m╧ë²a² = 2m╧Dzn²a²

12. Period of oscillation of mass joined to a spring

T = 2π \(\sqrt{\frac{m}{k}}\)
k → spring constant

13. Spring constant k

k ∝ \(\frac{1}{\ell}\)
If spring cut in two parts l₁ & l₂ then

k₁ = \(\frac{\ell}{\ell_{1}} \mathbf{k}=\frac{\ell_{1}+\ell_{2}}{\ell_{1}} \mathbf{k}p\), k₂ = \(\frac{\ell_{1}+\ell_{2}}{\ell_{2}} \mathrm{k}\)
If spring cut in to n equal parts then spring constant of each part will be k’ = nk

14. Springs in parallel

Effective force constant k’ = k₁ + k₂ + …….. + k_n
T’ = 2╧Ç \(\sqrt{\frac{m}{k^{\prime}}}\)


For n identical springs k’ = nk
T’ = \(\frac{T}{\sqrt{n}}\)
T → Time period due to one spring only Springs in series

15. Spring in series

If effective force constant is k’ then
\(\frac{1}{k^{\prime}}=\frac{1}{k_{1}}+\frac{1}{k_{2}}+\ldots \ldots \ldots . .+\frac{1}{k_{n}}\)
For identical springs \(\frac{1}{k^{\prime}}=\frac{n}{k}\)
and T’ = T\(\sqrt{n}\)
For two springs \(\frac{1}{k^{\prime}}=\frac{1}{k_{1}}+\frac{1}{k_{2}}\)
or k’ = \(\frac{\mathrm{k}_{1} \mathrm{k}_{2}}{\mathrm{k}_{1}+\mathrm{k}_{2}}\)
T’ = \(2 \pi \sqrt{\frac{m\left(k_{1}+k_{2}\right)}{k_{1} k_{2}}}\)

16. Two masses connected by two ends of a spring

T = 2π\(\sqrt{\frac{\mu}{\mathrm{k}}}\)
where ┬╡ = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\) = Reduced mass.

17. Oscillations of a liquid in a U-tube

T = 2π \(\sqrt{\frac{h}{g}}\) h is vertical height of liquid column.

18. Oscillations of a floating body

Cylindrical body
T =2 π\(\sqrt{\frac{m}{a d g}}\)
T = 2π\(\sqrt{\frac{h}{g}}\),
m = mass of immersed part
Rectangular body T = 2π\(\sqrt{\frac{h}{g}}\)
d → density of liquid,
a → cross sectional area,
m → mass of body.
h → A height of block or cylinder inside the liquid

19. Period of a simple pendulum

T = 2π\(\sqrt{\frac{\ell}{g}}\) (l << R)
T = 2π\(\sqrt{\frac{\mathrm{R}}{\mathrm{g}\left(1+\frac{\mathrm{R}}{\ell}\right)}}\)
when l is large, of the order of R.
T = 2π\(\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}\)
when l → ∞, T = 84.4 minute
Motion of a pendulum is simple harmonic only when the maximum angular displacement ╬╕₀ is small.

20. Second pendulum

T = 2 second; l = 96 cm.

21. Conical pendulum

T = 2π\(\sqrt{\frac{r}{g \tan \theta}}=\sqrt{\frac{L \cos \theta}{g}}\)
= 2π\(\left[\frac{\sqrt{L^{2}-r^{2}}}{g}\right]^{1 / 2}=2 \pi \sqrt{\frac{\ell}{g}}\)

22. Period in a reference frame moving with acceleration ΓÇÿaΓÇÖ in a horizontal plane

T = 2π\(\left[\frac{\ell}{\sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}}\right]^{1 / 2}\)

23. Oscillation of pendulum in a lift

Lift moving up with an acceleration a,
T = 2π\(\sqrt{\frac{\ell}{\mathrm{g}+\mathrm{a}}}\)

Lift moving down with an acceleration a,
T = 2π\(\sqrt{\frac{\ell}{g-a}}\)

Lift falling freely (a = g), T = ∞

Lift moving with a uniform speed (a = 0),
T = 2π\(\sqrt{\frac{\ell}{\mathrm{g}}}\)

24. Oscillations of pendulum having bob of density p in a liquid of density d

T = 2π\(\sqrt{\frac{\ell}{g\left(1-\frac{d}{\rho}\right)}}=2 \pi \sqrt{\frac{\ell}{g\left(1-\frac{1}{n}\right)}}\)
where \(\frac{\mathrm{d}}{\rho}=\frac{1}{\mathrm{n}}\) or n = \(\frac{\rho}{\mathrm{d}}\)

25. Vertical oscillations of a body suspended from a wire of YoungΓÇÖs modulus Y

T = 2π\(\sqrt{\frac{m L}{Y A}}\)
A → Cross sectional Area of wire
m → mass of the body
L → Length of the wire
Y → Young’s modulus of elasticity

26. Oscillation of a small ball placed in a concave glass

T = 2π\(\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}\)

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