1. ⇒ (MHT CET 2023 12th May Evening Shift )

A particle starts from mean position and performs S.H.M. with period 4 second. At what time its kinetic energy is $50 %$ of total energy?

$(\cos 45^{\circ} = \frac{1}{\sqrt{2}})$

A. 0.1 s

B. 0.2 s

C. 0.4 s

D. 0.5 s

Correct Option is (D)

The total energy is given by $T \cdot E = \frac{1}{2} {kA}^{2}$

Kinetic energy is given by $K \cdot E = \frac{1}{2} k (A^{2} - x^{2})$

$\begin{array}{ll} K.E = \frac{1}{2} P . E ...(given) \\ ∴ & \frac{1}{2} k (A^{2} - x^{2}) = \frac{1}{2} (\frac{1}{2} k A^{2}) \\ ∴ & (A^{2} - x^{2}) = \frac{1}{2} A^{2} \\ ∴ & (A^{2} - x^{2}) = \frac{1}{2} A^{2} \\ ∴ & x = \frac{A}{\sqrt{2}} \end{array}$

The equation of displacement in SHM is

$\begin{aligned} x = A \sin \frac{2 π t}{T} \\ ∴ \frac{A}{\sqrt{2}} = A \sin \frac{2 π t}{4} . . . . (∵ T = 4 sec) \\ ∴ \frac{1}{\sqrt{2}} = \sin \frac{π t}{2} \\ ∴ \frac{π t}{2} = \sin^{- 1} \frac{1}{\sqrt{2}} \\ \frac{π t}{2} = \frac{π}{4} \\ ∴ t = 0.5 s \end{aligned}$

2. ⇒ (MHT CET 2023 11th May Morning Shift )

For a particle executing S.H.M., its potential energy is 8 times its kinetic energy at certain displacement ' $x$ ' from the mean position. If ' $A$ ' is the amplitude of S.H.M the value of ' $x$ ' is

A. $\frac{A \sqrt{2}}{3}$

B. $A \sqrt{3}$

C. $\frac{2 \sqrt{2} A}{3}$

D. $\frac{A}{\sqrt{2}}$

Correct Option is (C)

Potential energy: $U = \frac{1}{2} m ω^{2} x^{2}$ and

Kinetic energy: $K = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$

Given: Potential energy $= 8 \times$ Kinetic energy

$\begin{aligned} ∴ U = 8 K \\ \frac{1}{2} m ω^{2} x^{2} = 8 \times \frac{1}{2} m ω^{2} (A^{2} - x^{2}) \\ x^{2} = 8 A^{2} - 8 x^{2} \\ 9 x^{2} = 8 A^{2} \\ x^{2} = \frac{8 A^{2}}{9} \\ x = \frac{2 \sqrt{2} A}{3} \end{aligned}$

3. ⇒ (MHT CET 2023 10th May Evening Shift )

A body is executing a linear S.H.M. Its potential energies at the displacement ' $x$ ' and ' $y$ ' are ' $E_{1}$ ' and ' $E_{2}$ ' respectively. Its potential energy at displacement $(x + y)$ will be

A. $E_{1} + E_{2}$

B. ${(\sqrt{E_{1}} + \sqrt{E_{2}})}^{2}$

C. $E_{1} - E_{2}$

D. ${(\sqrt{E_{2}} - \sqrt{E_{1}})}^{2}$

Correct Option is (B)

We know,

Potential Energy $E_{P} = \frac{1}{2} {Kx}^{2}$

$∴ E_{1} = \frac{1}{2} {Kx}^{2} \Rightarrow x = \sqrt{\frac{2 E_{1}}{K}}$ .... (i)

and

$E_{2} = \frac{1}{2} K y^{2} \Rightarrow y = \sqrt{\frac{2 E_{2}}{K}}$ ..... (ii)

Given, total displacement $= (x + y)$

$∴$ Potential energy at displacement $(x + y)$ ,

$E$ is $\frac{1}{2} K (x + y)^{2}$

$\begin{aligned} = \frac{1}{2} K {(\sqrt{\frac{2 E_{1}}{K}} + \sqrt{\frac{2 E_{2}}{K}})}^{2} \\ = \frac{1}{2} K [\frac{2 E_{1}}{K} + \frac{2 E_{2}}{K} + 2 (\sqrt{\frac{E_{1}}{K}}) (\sqrt{\frac{2 E_{2}}{K}})] \\ = {(\sqrt{E_{1}} + \sqrt{E_{2}})}^{2} = (E_{1} + E_{2} + 2 \sqrt{E_{1} E_{2}}) \end{aligned}$

4. ⇒ (MHT CET 2023 9th May Evening Shift )

A particle is vibrating in S.H.M. with an amplitude of $4 cm$ . At what displacement from the equilibrium position is its energy half potential and half kinetic?

A. $1 cm$

B. $\sqrt{2} cm$

C. $2 cm$

D. $2 \sqrt{2} cm$

Correct Option is (D)

$\begin{aligned} K.E & = \frac{1}{2} m ω^{2} (A^{2} - x^{2}) and \\ P.E & = \frac{1}{2} m ω^{2} x^{2} \\ ∴ T.E & = K.E + P.E \\ = \frac{1}{2} m ω^{2} (A^{2} - x^{2}) + \frac{1}{2} m ω^{2} x^{2} \\ = \frac{1}{2} m ω^{2} A^{2} \\ Given: K.E = P.E \\ ∴ A^{2} & - x^{2} = x^{2} \\ ∴ & A^{2} = 2 x^{2} \\ ∴ x & = \sqrt{\frac{A^{2}}{2}} = \frac{A}{\sqrt{2}} \\ ∴ x & = \frac{4}{\sqrt{2}} = 2 \sqrt{2} c m \end{aligned}$

5. ⇒ (MHT CET 2021 21th September Evening Shift )

A child is sitting on a swing which performs S.H.M. It has minimum and maximum heights from ground $0.75 cm$ and $2 m$ respectively. Its maximum speed will be $[g = 10 \frac{m}{s^{2}}]$

A. $\sqrt{1.25} m / s$

B. $\sqrt{12.5} m / s$

C. $5 m / s$

D. $25 m / s$

Correct Option is (C)

Maximum potential energy is attained at the highest point which gets converted into kinetic energy at the lowest point.

$\begin{aligned} h = 2 - 0.75 = 1.25 m \\ \frac{1}{2} {mv}^{2} = mgh \\ ∴ v^{2} = 2 gh = 2 \times 10 \times 1.25 = 25 \\ ∴ v = 5 m / s \end{aligned}$

⇒Oscillations