NEET Physics Oscillation Notes

Oscillation Displacement

Displacement in SHM can be represented by,

y = a sin cot → (1)

Where ‘y’ is instantaneous displacement, ‘a’ is amplitude, and ‘Q’ is angular frequency.

Simple Harmonic Motion (SHM) NEET Notes

Oscillation Velocity

Velocity of a particle in SHM

v = \(\frac{d y}{d t}=\frac{d}{d t}(a \sin \omega t) \Rightarrow v=a \omega \cos \omega t\) → (2)

v = \(a \omega\left[1-\sin ^2 \omega t\right]^{\frac{1}{2}}\)

v = \(\omega \sqrt{a^2-a^2 \sin ^2 \omega t}\)

v = \(\omega \sqrt{a^2-y^2}\)

⇒ \(v^2=\omega^2\left(a^2-y^2\right)\)

⇒ \(\frac{v^2}{\omega^2}=a^2-y^2 \Rightarrow \frac{v^2}{a^2 \omega^2}=1-\frac{y^2}{a^2}\)

⇒ \(\frac{v^2}{a^2 \omega^2}+\frac{y^2}{a^2}=1\)

Which is an equation of ellipse. This means, the graph of v versus y in SHM is an ellipse.

Read And Learn More: NEET Physics Notes

Oscillation Acceleration

Acceleration of SHM is given by,

NEET Study Material for Oscillations Chapter

A = \(\frac{d v}{d t}=\frac{d}{d t} a \omega \cos \omega t\)

A = \(-a \omega^2 \sin \omega t\) → (3)

A = \(-\omega^2 \mathrm{y}\)→(4)

Note: \(v_{\max }=a \omega ; A_{\max }=-\omega^2 a\)

NEET Physics Oscillations Notes

Oscillation Potential Energy

The potential energy of a particle in SHM is,

U = \(\frac{1}{2} \mathrm{ky}^2\)

Where, k = \(m \omega^2\)

Oscillation Kinetic Energy

The kinetic energy of a particle in SHM is,

K = \(\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(a^2-y^2\right)\)

∴ Total energy is, E = U + K

= \(\frac{1}{2}\)ky² + \(\frac{1}{2}\)k(a²-y²)

= \(\frac{1}{2}\)ky² + \(\frac{1}{2}\)ka² – \(\frac{1}{2}\)ky²

E = \(\frac{1}{2}\)ka²

NEET Physics Oscillations Important Formulas

Note:

• From the above equation, we see that energy is directly proportional to the square of the amplitude
• Also, intensity (I) is directly proportional to the square of the amplitude

I = \(\frac{E}{A t} \ E \propto a^2\)

∴ I \(\propto a^2\)

Time of oscillation of a simple pendulum (for small oscillations) is given by

T = \(2 \pi \sqrt{\frac{\ell}{g}}\)

Period of oscillation of a spring mass system is given by,

⇒ \(2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}\)

Where k is the spring constant.

If a spring of spring constant k is cut into two halves, each part will have a spring constant equal to 2k.

Oscillation

Best Short Notes for Oscillations NEET

The above point can be understood easily by using,

F = kx and Y = \(\frac{F}{A \Delta}\) equation

If a spring of spring constant k is cut into n equal halves, the each part will have a spring constant equal to nk.

NEET Physics Oscillations MCQs with Solutions

The series combination of springs

⇒ \(\frac{1}{\mathrm{~K}_{\mathrm{eff}}}=\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2}+\frac{1}{\mathrm{k}_3}+\ldots .+\frac{1}{\mathrm{k}_{\mathrm{a}}}\)

If there are two springs, then

⇒ \(\frac{1}{\mathrm{~K}_{\text {efe }}}=\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2} \)

⇒ \(\frac{1}{\mathrm{~K}_{\text {eff }}}=\frac{\mathrm{k}_1+\mathrm{k}_2}{\mathrm{k}_1 \mathrm{k}_2}\)

⇒ \(\mathrm{~K}_{\text {eff }}=\frac{\mathrm{k}_1 \mathrm{k}_2}{\mathrm{k}_1+\mathrm{k}_2}\)

When ‘n’ springs of equal spring constants k are connected in series, then K_eff = nk

Oscillations NEET Important Questions and Answers

Parallel combination of springs: When ‘n’ springs of spring constants kWhen ‘n’ springs of spring constants k₁, k_2, and so on are connected in series, then, k₂, ….k_n are connected in parallel, then,

K_eff = k₂ + k₂ + ….. + k_n

When ‘n’ springs of equal spring constants k are connected in parallel, then K_eff = nk