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 4

Q58 jee_main_2024_29_jan_morning Energy in Simple Harmonic Motion

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is

x8

, where

x =

________.

Numerical Answer. Answer: 9 to 9

Solution

Related Formula

Total energy ( $E$ $E$) in simple harmonic motion is:

E = (1)/(2) k A²

Potential energy ( $U$ $U$) at displacement $y$ $y$ is:

U = (1)/(2) k y²

Kinetic energy ( $KE$ $KE$) is the remaining energy:

$KE = E - U$ $KE = E - U$

Core Logic

Given the displacement is one-third of the amplitude:

y = (A)/(3)

Substituting this displacement into the potential energy expression:

U = (1)/(2) k ((A)/(3))² = (1)/(9) ( (1)/(2) k A² ) = (E)/(9)

Step 1: Calculate Kinetic Energy The kinetic energy is:

KE = E - (E)/(9) = (8E)/(9)

Step 2: Find the Energy Ratio The ratio of total energy to kinetic energy is:

(E)/(KE) = (E)/((8E)/(9)) = (9)/(8)

Comparing this with the given target form$ Comparing this with the given target form $\frac{x}{8}: $: $x = 9 Therefore, the value of$ Therefore, the value of $xis$ is $9. Pattern Recognition Since$. Pattern Recognition Since $U \propto y^2, if displacement scales by a fraction$, if displacement scales by a fraction $\frac{1}{n}, potential energy scales instantly by$, potential energy scales instantly by $\frac{1}{n^2}. The remaining kinetic energy is naturally$. The remaining kinetic energy is naturally $1 - \frac{1}{n^2}, giving an immediate shortcut for the total-to-kinetic ratio:$, giving an immediate shortcut for the total-to-kinetic ratio: $\frac{n^2}{n^2 - 1}$. Chapter Mix Class 11 Physics: Oscillations

Q58 jee_main_2024_30_january_evening Simple Pendulum Time Period

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is

4 ~m

, then the time period of small oscillations will be ________

s

. [take

g = π² ~m/s²

]

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

g' = (GM)/((R+h)²) = g((R)/(R+h))²

T = 2π \sqrt((l)/(g'))

Core Logic

The distance of the pendulum from the earth's surface is $h = R$ $h = R$. The acceleration due to gravity at this height $g'$ $g'$ is:

g' = g((R)/(R+R))² = g((1)/(2))² = (g)/(4)

Step 1: Calculate Time Period

The time period of the simple pendulum is:

T = 2π \sqrt((l)/(g'))

Substitute $l = 4 ~m$ $l = 4 \mathrm{~m}$ and $g' = (g)/(4)$ $g' = \frac{g}{4}$:

T = 2π \sqrt((4)/(g/4)) = 2π \sqrt((16)/(g))

Given $g = π²$ $g = \pi^2$:

T = 2π \sqrt((16)/(π²)) = 2π ((4)/(π)) = 8 ~s

Pattern Recognition

At $h=R$ $h=R$, $g$ $g$ drops to $g/4$ $g/4$. Thus, the time period doubles compared to its surface value.

Chapter Mix

Class 11 Physics: Oscillations Class 11 Physics: Gravitation

Q59 jee_main_2024_31_jan_evening Spring Mass System

The time period of simple harmonic motion of mass M in the given figure is

π \sqrt((α M)/(5K))

, where the value of

α

Spring Mass System diagram for Q59 - JEE Main 2024 Evening — The image shows a combination of springs attached to a mass. Springs of stiffness 2k and k are connected in parallel, which in turn are connected in series with a spring of stiffness k.

Numerical Answer. Answer: 12 to 12

Solution

Related Formula

Springs in parallel: $kₚ = k₁ + k₂$ $k_p = k_1 + k_2$ Springs in series:

(1)/(kₛ) = (1)/(kₚ) + (1)/(k₃)

Time period:

T = 2π Mkeq

Core Logic

Evaluate the equivalent spring constant of the given setup by first resolving the parallel configuration, then the series configuration.

Step 1: Equivalent Spring Constant

Based on the standard layout of the problem (as implied by the solution text), let's assume one section is $2k$ $2k$ and $k$ $k$ in parallel... Wait, the solution implies $(2k \cdot k)/(3k) + k$ $\frac{2k \cdot k}{3k} + k$, which represents a series combination of $2k$ $2k$ and $k$ $k$, followed by a parallel combination with another $k$ $k$. Let's trust the official PDF's algebraic step: $keq = (2k \cdot k)/(3k) + k$ $k_{eq} = \frac{2k \cdot k}{3k} + k$. This evaluates to:

keq = (2k)/(3) + k = (5k)/(3)

Step 2: Time Period Calculation

T = 2π Mkeq

T = 2π \sqrt((M)/(5k/3)) = 2π \sqrt((3M)/(5k))

Step 3: Matching the Format

Bring the 2 inside the square root as 4:

T = π \sqrt((4 \times 3M)/(5k)) = π \sqrt((12M)/(5k))

Comparing with $π \sqrt((α M)/(5k))$ $\pi \sqrt{\frac{\alpha M}{5k}}$: $α = 12$ $\alpha = 12$

Pattern Recognition

Whenever an external coefficient like 2 (from $2π$ $2\pi$) is pushed inside the square root to match a format, it squares. Don't forget $2 \to 4$ $2 \to 4$, turning $3M$ $3M$ into $12M$ $12M$.

Chapter Mix

Class 11 Physics: Oscillations

Q59 jee_main_2024_31_jan_morning Velocity In SHM

A particle performs simple harmonic motion with amplitude

A

. Its speed is increased to three times at an instant when its displacement is

(2A)/(3)

. The new amplitude of motion is

(nA)/(3)

. The value of

n

is ______.

Numerical Answer. Answer: 7 to 7

Solution

Related Formula

v = ω \sqrt(A² - x²)

Core Logic

At displacement $x = (2A)/(3)$ $x = \frac{2A}{3}$, the initial velocity is:

v = ω \sqrt(A² - ((2A)/(3))²)

v = ω \sqrt(A² - (4A²)/(9)) = ω \sqrt((5A²)/(9))

v = \sqrt(5)Aω3

Step 2: Velocity Tripled

The speed is now increased to $v' = 3v$ $v' = 3v$ at the same position $x = (2A)/(3)$ $x = \frac{2A}{3}$.

v' = 3 ( \sqrt(5)Aω3) = \sqrt(5)Aω

This new velocity corresponds to a new amplitude $A'$ $A'$:

v' = ω \sqrt((A')² - x²)

\sqrt(5)Aω = ω \sqrt((A')² - ((2A)/(3))²)

Step 3: Finding new Amplitude

Squaring both sides:

5A² = (A')² - (4A²)/(9)

(A')² = 5A² + (4A²)/(9) = (45A² + 4A²)/(9)

(A')² = (49A²)/(9)

A' = (7A)/(3)

Comparing with $(nA)/(3)$ $\frac{nA}{3}$, we get $n = 7$ $n = 7$.

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 4

Solution

Related Formula

Core Logic

Step 1: Calculate Kinetic Energy

Step 1: Calculate Kinetic Energy

Step 2: Find the Energy Ratio

Step 2: Find the Energy Ratio

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Time Period

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Equivalent Spring Constant

Step 2: Time Period Calculation

Step 3: Matching the Format

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Velocity Tripled

Step 3: Finding new Amplitude

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.