A simple harmonic oscillator has an amplitude A and time period 6pi second. Assuming the oscillation starts from its…

Q33 jee_main_2026_21_jan_evening Simple Harmonic Motion

The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of

176 rad/s

. The frequency of this simple harmonic oscillator is ________ Hz.

[take π = (22)/(7)]

A. $14$ $14$
B. $88$ $88$
C. $28$ $28$
D. $176$ $176$

Solution

Related Formula

ωk = 2π fk

foscillator = (fk)/(2)

Core Logic

For a simple harmonic oscillator with an angular frequency $ω$ $\omega$, the kinetic energy oscillates at double the angular frequency ( $2ω$ $2\omega$).

The given angular frequency of kinetic energy oscillation is $ωk = 176 rad/s$ $\omega_k = 176 \text{ rad/s}$.

Step 1: Finding Kinetic Energy Frequency

First, find the frequency of kinetic energy oscillations ( $fk$ $f_k$):

fk = (ωk)/(2π) = (176)/(2 \times (22)/(7))

fk = (176 \times 7)/(44) = 4 \times 7 = 28 Hz

Step 2: Final Conclusion

Since kinetic energy oscillates at twice the frequency of the oscillator itself:

foscillator = (fk)/(2) = (28)/(2) = 14 Hz

Pattern Recognition

KE and PE always oscillate at twice the frequency ( $2ν$ $2\nu$) of the underlying SHM ( $v$ $v$) due to the $²(ω t)$ $\sin^2(\omega t)$ or $²(ω t)$ $\cos^2(\omega t)$ terms resolving to $(1 - (2ω t))/2$ $(1 - \cos(2\omega t))/2$.

Chapter Mix

Class 11 Physics: Oscillations

Q27 jee_main_2026_22_january_evening Simple Pendulum

Using a simple pendulum experiment g is determined by measuring its time period T. Which of the following plots represent the correct relation between the pendulum length L and time period T?

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

Solution

Related Formula

T = 2π \sqrt((L)/(g))

Core Logic

Squaring both sides of the simple pendulum equation:

T² = (4π² L)/(g)

Inverting $T²$ $T^2$ to express $1/T²$ $1/T^2$ as a function of length $L$ $L$:

(1)/(T²) = (g)/(4π² L)

This demonstrates that $(1)/(T²) \propto (1)/(L)$ $\frac{1}{T^2} \propto \frac{1}{L}$. The relationship between $1/T²$ $1/T^2$ and $L$ $L$ represents a rectangular hyperbola curve decreasing asymptotically.

Simple Pendulum 1/T^2 vs L graph for Q27 - JEE Main 2026 Evening

Step 1: Final Conclusion

Plot (2) correctly depicts the rectangular hyperbola curve for $(1)/(T²)$ $\frac{1}{T^2}$ versus $L$ $L$.

Pattern Recognition

Sees: Graph of $1/T²$ $1/T^2$ vs $L$ $L$. Formula check: $T \propto \sqrt(L) T² \propto L 1/T² \propto 1/L$ , which is inversely proportional (hyperbolic curve).

Chapter Mix

Class 11 Physics: Simple Harmonic Motion

Q31 jee_main_2026_23_january_morning Simple Pendulum

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.]

A. $120$ $120$
B. $0.75$ $0.75$
C. $7.5$ $7.5$
D. $15$ $15$

Solution

Related Formula

T = 2π \sqrt((l)/(g))

T \propto \sqrt(l)

Core Logic

For a simple pendulum, the period is proportional to the square root of the length. If the frequency is doubled, the period is halved, which requires the length to become one-fourth of its original value.

Step 1: Find Old and New Period

Initial time period $T₁ = (10)/(20) = 0.5 s$ $T_{1} = \frac{10}{20} = 0.5 \text{ s}$ New time period $T₂ = (10)/(40) = 0.25 s$ So, $T₂ = T₁2$ $T_{2} = \frac{T_{1}}{2}$

Step 2: Calculate New Length

T₁T₂ = l₁l₂

2 = 30l₂

4 = 30l₂

l₂ = (30)/(4) = 7.5 cm

Pattern Recognition

Sees: "Oscillations doubled in same time" → Frequency is doubled → Time period is halved. Since $T \propto \sqrt(l)$ $T \propto \sqrt{l}$, $l$ $l$ must be reduced to $(1)/(4)$ $\frac{1}{4}$ of the original.

Chapter Mix

Class 11 Physics: Oscillations

Q28 jee_main_2026_24_january_morning Simple Harmonic Motion

A cylindrical block of mass M and area of cross section A is floating in a liquid of density

ρ

and with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation is ____.

A. $2π\sqrt((M)/(ρ Ag))$ $2\pi\sqrt{\frac{M}{\rho Ag}}$
B. $π 2M$
C. $π Mg$
D. $2π \sqrt((ρ A)/(Mg))$ $2\pi \sqrt{\frac{\rho A}{Mg}}$

Solution

Related Formula

T = 2π Mkeffective

Fbuoyancy = ρ V g

Core Logic

At equilibrium, the buoyant force balances the weight: $ρ Ahg = Mg$ $\rho Ahg = Mg$ After displacing the block downward by a small distance $x$ $x$, the net restoring force is:

Ma = -ρ A(h+x)g + Mg

Ma = -ρ Ahg - ρ Axg + Mg

Since $ρ Ahg = Mg$ $\rho Ahg = Mg$, this simplifies to: $Ma = -ρ Axg$ $Ma = -\rho Axg$

a = ((-ρ Ag)/(M)) x

Step 1: Compare with Standard SHM Equation

Comparing with $a = -ω² x$ $a = -\omega^2 x$, we get:

ω = \sqrt((ρ Ag)/(M))

The time period $T$ $T$ is:

T = (2π)/(ω) = 2π \sqrt((M)/(ρ Ag))

Pattern Recognition

For a floating body of uniform cross-section $A$ $A$, the restoring force constant is simply $k = ρ A g$ $k = \rho A g$. The time period is immediately $T = 2π \sqrt(m/k)$ $T = 2\pi \sqrt{m/k}$.

Chapter Mix

Class 11 Physics: Oscillations Class 11 Physics: Mechanical Properties of Fluids

Q46 jee_main_2026_28_january_morning Energy in SHM

The displacement of a particle, executing simple harmonic motion with time period T, is expressed as

x(t) = A ω t

, where A is the amplitude. The maximum value of potential energy of this oscillator is found at

t = T/2β

. The value of

β

is ____.

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

PE = (1)/(2) k x²

$PEmax$ $PE_{\text{max}}$ occurs when $x = \pm A$ $x = \pm A$.

Core Logic

Potential energy is maximum at the extreme positions. From $x(t) = A (ω t)$ $x(t) = A \sin(\omega t)$, the particle starts at the mean position ( $x=0$ $x=0$ at $t=0$ $t=0$) and reaches the extreme position ( $x=A$ $x=A$) for the first time at $t = T/4$ $t = T/4$.

Step 1: Finding beta

Given time for maximum PE is $t = (T)/(4)$ $t = \frac{T}{4}$. We are given $t = (T)/(2β)$ $t = \frac{T}{2\beta}$. Therefore, $(T)/(4) = (T)/(2β) \Rightarrow 2β = 4 \Rightarrow β = 2$ .

Pattern Recognition

Sine function SHM starts at mean, hits extreme at $T/4$ $T/4$, back to mean at $T/2$ $T/2$, negative extreme at $3T/4$ $3T/4$, back to mean at $T$ $T$. PE is max at $T/4$ $T/4$ and $3T/4$ $3T/4$.

Chapter Mix

Class 11 Physics: Oscillations

A simple harmonic oscillator has an amplitude A$A$ and time period 6π$6\pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from x = A$x = A$ to x = √(3)2A$x = \frac{\sqrt{3}}{2}A$ will be (π)/(x)$\frac{\pi}{x}$ s, where x =$x =$ ________.

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate the Time Difference

Pattern Recognition

Chapter Mix

More Oscillations Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Finding Kinetic Energy Frequency

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find Old and New Period

Step 2: Calculate New Length

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compare with Standard SHM Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding beta

Pattern Recognition

Chapter Mix

More Oscillations Questions — jee_main_2024_29_january_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back