A particle oscillates along the x-axis according to the law, x(t) = x0 2((t/2) ight) where x0 = 1 , m . The kinetic energy (K) of the particle as a function of x is correctly represented by the graph.

Given x0 = 1 m, the position simplifies to: Differentiating x(t) to find velocity v: Kinetic energy K is proportional to v2: Substituting t = 1 - 2x: This is an inverted parabola (K propto x - x2) passing through x=0 and x=1, reaching a peak at x = 1/2. This matches Graph (1). The position oscillates purely within the boundary interval [0, 1]. Kinetic energy peaks perfectly at the equilibrium midpoint x = 1/2 where speed is maximum. Class 11 Physics: Oscillations

A particle oscillates along the x-axis according to the law, x(t) = x0 2((t/2) ight) where x0 = 1 , m . The kinetic energy (K) of the particle as a function of x is correctly represented by the graph.

Given x0 = 1 m, the position simplifies to: Differentiating x(t) to find velocity v: Kinetic energy K is proportional to v2: Substituting t = 1 - 2x: This is an inverted parabola (K propto x - x2) passing through x=0 and x=1, reaching a peak at x = 1/2. This matches Graph (1). The position oscillates purely within the boundary interval [0, 1]. Kinetic energy peaks perfectly at the equilibrium midpoint x = 1/2 where speed is maximum. Class 11 Physics: Oscillations

A particle oscillates along the x-axis according to the law, x(t) = x0 2((t/2) ight) where x0 = 1 , m . The kinetic energy (K) of the particle as a function of x is correctly represented by the graph.

Given x0 = 1 m, the position simplifies to: Differentiating x(t) to find velocity v: Kinetic energy K is proportional to v2: Substituting t = 1 - 2x: This is an inverted parabola (K propto x - x2) passing through x=0 and x=1, reaching a peak at x = 1/2. This matches Graph (1). The position oscillates purely within the boundary interval [0, 1]. Kinetic energy peaks perfectly at the equilibrium midpoint x = 1/2 where speed is maximum. Class 11 Physics: Oscillations

A particle oscillates along the x-axis according to the law, x(t) = x0 2((t/2) ight) where x0 = 1 , m . The kinetic energy (K) of the particle as a function of x is correctly represented by the graph.

Given x0 = 1 m, the position simplifies to: Differentiating x(t) to find velocity v: Kinetic energy K is proportional to v2: Substituting t = 1 - 2x: This is an inverted parabola (K propto x - x2) passing through x=0 and x=1, reaching a peak at x = 1/2. This matches Graph (1). The position oscillates purely within the boundary interval [0, 1]. Kinetic energy peaks perfectly at the equilibrium midpoint x = 1/2 where speed is maximum. Class 11 Physics: Oscillations

A particle oscillates along the x-axis according to the law, x(t) = x0 2((t/2) ight) where x0 = 1 , m . The kinetic…

More Oscillations Previous-Year Questions — Page 2

Q42 jee_main_2026_28_january_evening Two Body Oscillator

As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses

1 kg

and

0.2 kg

with a separation more than spring natural length and are released. Assuming the horizontal surface to be frictionless, the angular frequency (in SI unit) of the system is :

Two Body Oscillator diagram for Q42 - JEE Main 2026 Evening — Two masses attached to ends of a spring on a horizontal table.

A. $30$ $30$
B. $27$ $27$
C. $20$ $20$
D. $5$ $5$

Solution

Related Formula

ω = \sqrt((k)/(μ))

where $μ$ $\mu$ is the reduced mass of the system given by:

μ = (m₁ m₂)/(m₁ + m₂)

Step 1: Calculate Reduced Mass

Given $m₁ = 1 kg$ $m_1 = 1 \text{ kg}$ and $m₂ = 0.2 kg$ $m_2 = 0.2 \text{ kg}$.

μ = (1 \times 0.2)/(1 + 0.2) = (0.2)/(1.2)

μ = (1)/(6) kg

Step 2: Calculate Angular Frequency

Given spring constant $k = 150 N/m$ $k = 150 \text{ N/m}$.

ω = \sqrt((150)/(1/6)) = \sqrt(150 \times 6) = \sqrt(900)

ω = 30 rad/s

Pattern Recognition

For any two-body spring system free of external forces, always substitute the individual mass $m$ $m$ with the reduced mass $μ = (m₁ m₂)/(m₁ + m₂)$ $\mu = \frac{m_1 m_2}{m_1 + m_2}$ in the standard oscillator formula $ω = \sqrt(k/m)$ $\omega = \sqrt{k/m}$.

Chapter Mix

Class 11 Physics: Oscillations

Q jee_main_2025_02_april_morning Superposition of SHMs

A particle is subjected to two simple harmonic motions as:

x₁ = \sqrt(7) 5t~cm

and

x₂ = 2\sqrt(7) (5t + (π)/(3))~cm

where

x

is displacement and

t

is time in seconds. The maximum acceleration of the particle is

x \times 10⁻²~ms⁻²

. The value of

x

is:

A. 175
B. $25\sqrt(7)$ $25\sqrt{7}$
C. $5\sqrt(7)$ $5\sqrt{7}$
D. 125

Solution

Related Formula

A = \sqrt(A₁² + A₂² + 2 A₁ A₂ φ)

a = ω² A

Core Logic

Both SHMs share the same frequency $ω = 5~rad/s$ $\omega = 5\mathrm{~rad/s}$ with phase difference $φ = (π)/(3)$ $\phi = \frac{\pi}{3}$. Amplitudes: $A₁ = \sqrt(7)~cm$ $A_1 = \sqrt{7}\mathrm{~cm}$, $A₂ = 2\sqrt(7)~cm$ $A_2 = 2\sqrt{7}\mathrm{~cm}$.

The combined amplitude is:

A = (\sqrt(7))² + (2\sqrt(7))² + 2(\sqrt(7))(2\sqrt(7)) ((π)/(3))

A = \sqrt(7 + 28 + 2(7)(2)((1)/(2))) = \sqrt(35 + 14) = \sqrt(49) = 7~cm

Converting to meters:

A = 0.07~m

The maximum acceleration is:

a = ω² A = (5)² \times 0.07 = 25 \times 0.07 = 1.75~ms⁻²

Equating to $x \times 10⁻²~ms⁻²$ $x \times 10^{-2}\mathrm{~ms}^{-2}$:

1.75 = x \times 10⁻² x = 175

Step 1: Final Conclusion

The value of $x$ $x$ is 175.

Pattern Recognition

When superposing two SHMs of identical frequency, use the standard phasor addition formula to obtain the combined amplitude $A$ $A$. Then compute maximum velocity $v = ω A$ $v_{\max} = \omega A$ or maximum acceleration $a = ω² A$ $a_{\max} = \omega^2 A$ directly.

Chapter Mix

Class 11 Physics: Simple Harmonic Motion

Q10 jee_main_2025_29_jan_evening Energy in Simple Harmonic Motion

Two bodies A and B of equal mass are suspended from two massless springs of spring constant

k₁

and

k₂

, respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is:

A. $\sqrt((k₁)/(k₂))$ $\sqrt{\frac{k_1}{k_2}}$
B. $k₁k₂$ $\frac{\mathrm{k}_1}{\mathrm{k}_2}$
C. $k₂k₁$ $\frac{\mathrm{k}_2}{\mathrm{k}_1}$
D. $k₂k₁$ $\sqrt{\frac{k_{2}}{k_{1}}}$

Solution

Related Formula

v = Aω

ω = \sqrt((k)/(m))

Core Logic

The maximum velocity in vertical SHM happens at the mean equilibrium position and is given by $v = Aω$ $v_{\max} = A\omega$.

Given $mA = mB = m$ $m_A = m_B = m$ and $AA = AB = A$ $A_A = A_B = A$:

vA = A ω₁ = A \sqrt((k₁)/(m))

vB = A ω₂ = A \sqrt((k₂)/(m))

Taking the ratio:

(vA)/(vB) = (ω₁)/(ω₂) = \sqrt((k₁)/(m))\sqrt((k₂)/(m)) = \sqrt((k₁)/(k₂))

Pattern Recognition

Since maximum velocity is directly proportional to angular frequency $ω$ $\omega$ for equal amplitudes, and $ω \propto \sqrt(k)$ $\omega \propto \sqrt{k}$, the velocity ratio directly yields the square root of the spring constant ratio.

Chapter Mix

Class 11 Physics: Oscillations

Q jee_main_2025_04_april_morning Simple Pendulum

Two simple pendulums having lengths

l₁

and

l₂

with negligible string mass undergo angular displacements

θ₁

and

θ₂

, from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?

A. $θ₁l₂² = θ₂l₁²$
B. $θ₁l₁ = θ₂l₂$ $\theta_{1}l_{1} = \theta_{2}l_{2}$
C. $θ₁l₁² = θ₂l₂²$
D. $θ₁l₂ = θ₂l₁$ $\theta_{1}l_{2} = \theta_{2}l_{1}$

Solution

Related Formula

Angular acceleration definition for a simple pendulum swinging at small angle $θ$ $\theta$:

α = -ω² θ

where:

ω = \sqrt((g)/(l)) ω² = (g)/(l)

Hence, the magnitude of angular acceleration is:

α = (g)/(l)θ

Core Logic

Given that angular accelerations are exactly identical in magnitude ( $α₁ = α₂$ $\alpha_1 = \alpha_2$):

(g)/(l₁)θ₁ = (g)/(l₂)θ₂

Step 1: Simplify Expression

Cancelling out the constant gravitational acceleration $g$ $g$:

(θ₁)/(l₁) = (θ₂)/(l₂) θ₁ l₂ = θ₂ l₁

Pattern Recognition

Angular acceleration scales inversely with length for a fixed angular displacement ( $α \propto (θ)/(l)$ $\alpha \propto \frac{\theta}{l}$). To maintain identical angular accelerations, the ratio $(θ)/(l)$ $\frac{\theta}{l}$ must remain constant, yielding the cross-multiplication relation $θ₁ l₂ = θ₂ l₁$ $\theta_1 l_2 = \theta_2 l_1$.

Evaluation Rubric / Model Answer

Option D: $θ₁l₂ = θ₂l₁$ $\theta_{1}l_{2} = \theta_{2}l_{1}$

Chapter Mix

Class 11 Physics: Oscillations

A particle oscillates along the x-axis according to the law, x(t) = x₀ ²((t)/(2))$\mathrm{x}(t) = \mathrm{x}_{0} \sin^{2}\left(\frac{t}{2}\right)$ where x₀ = 1 m$x_{0} = 1 \, m$ . The kinetic energy (K) of the particle as a function of x is correctly represented by the graph.

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Oscillations Previous-Year Questions — Page 2

Solution

Related Formula

Step 1: Calculate Reduced Mass

Step 2: Calculate Angular Frequency

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify Expression

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

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A particle oscillates along the x-axis according to the law, $x(t) = x₀ ²((t)/(2))$ where $x₀ = 1 m$ $x_{0} = 1 \, m$ . The kinetic energy (K) of the particle as a function of x is correctly represented by the graph.