A simple pendulum of string length 30 cm performs 20 oscillations in 10s. The length of the string required for the…

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 simple pendulum of string length 30 cm performs 20 oscillations in 10s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is ____ cm. [Assume that the mass of the pendulum remains same.]

Solution & Explanation

Related Formula

Core Logic

Step 1: Find Old and New Period

Step 2: Calculate New Length

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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