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…

More Oscillations Previous-Year Questions — Page 2

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

Q19 jee_main_2025_24_jan_evening Simple Harmonic Motion Graphs

A particle oscillates along the x-axis according to the law,

x(t) = x₀ ²((t)/(2))

where

x₀ = 1 m

. The kinetic energy (K) of the particle as a function of x is correctly represented by the graph.

A. Graph (1)
B. Graph (2)
C. Graph (3)
D. Graph (4)

Solution

Related Formula

x(t) = x₀ ²((t)/(2)) = x₀ ((1 - t)/(2))

Core Logic

Given $x₀ = 1 m$ $x_0 = 1\ \mathrm{m}$, the position simplifies to:

x = (1 - t)/(2) 2x - 1 = - t t = 1 - 2x

Differentiating $x(t)$ $x(t)$ to find velocity $v$ $v$:

v = (dx)/(dt) = (1)/(2) t

Kinetic energy $K$ $K$ is proportional to $v²$ $v^2$:

K = (1)/(2)mv² = (1)/(2)m ((1)/(4) ² t) = (m)/(8)(1 - ² t)

Substituting $t = 1 - 2x$ $\cos t = 1 - 2x$:

K = (m)/(8)[1 - (1 - 2x)²] = (m)/(8)[1 - (1 - 4x + 4x²)] = (m)/(8)(4x - 4x²) = (m)/(2)(x - x²)

This is an inverted parabola ( $K \propto x - x²$ $K \propto x - x^2$) passing through $x=0$ $x=0$ and $x=1$ $x=1$, reaching a peak at $x = 1/2$ $x = 1/2$. This matches Graph (1).

Pattern Recognition

The position oscillates purely within the boundary interval $[0, 1]$ $[0, 1]$. Kinetic energy peaks perfectly at the equilibrium midpoint $x = 1/2$ $x = 1/2$ where speed is maximum.

Chapter Mix

Class 11 Physics: Oscillations

Solution & Explanation

Related Formula

Step 1: Calculate Reduced Mass

Step 2: Calculate Angular Frequency

Pattern Recognition

Chapter Mix

Reference Study Guides

More Oscillations Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

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Solution

Related Formula

Core Logic

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Simplify Expression

Pattern Recognition

Evaluation Rubric / Model Answer

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Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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