Match $List-I$ with $List-II$:
| | $List-I$ ($x-y$ graphs) | | $List-II$ (Situations) |
|---|---|---|---|
| $(a)$ | Damped oscillation graph | $(i)$ | Total mechanical energy is conserved |
| $(b)$ | Linear graph $y = -kx$ | $(ii)$ | Bob of a pendulum is oscillating under negligible air friction |
| $(c)$ | Simple harmonic motion graph | $(iii)$ | Restoring force of a spring |
| $(d)$ | Energy conservation graph ($K$.$E$. and $P$.$E$. curves) | $(iv)$ | Bob of a pendulum is oscillating along with air friction |
Choose the correct answer from the options given below:

  • A
    $(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)$
  • B
    $(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)$
  • C
    $(a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)$
  • D
    $(a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)$

Explore More

Similar Questions

The center of a disk of radius $r$ and mass $m$ is attached to a spring of spring constant $k$,inside a ring of radius $R > r$ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring,without slipping. The spring can only be stretched or compressed along the periphery of the ring,following Hooke's law. In equilibrium,the disk is at the bottom of the ring. Assuming small displacement of the disc,the time period of oscillation of the center of mass of the disk is written as $T = \frac{2 \pi}{\omega}$. The correct expression for $\omega$ is ($g$ is the acceleration due to gravity):

$A$ block of mass $(10 \alpha) \text{ g}$,where $\alpha$ is a constant,is moving with velocity $3 \text{ m/s}$ to the right. It collides inelastically with a block on the right of mass $10 \text{ g}$ and sticks to it. The right block is connected to three springs as shown in the figure. The spring constant of each spring is $k = 2 \text{ N/m}$. If the amplitude of the resulting simple harmonic motion is $A = \frac{1}{2\sqrt{2}} \text{ m}$,then the value of $\alpha$ is:

As shown in the figures,a uniform rod $OO^{\prime}$ of length $l$ is hinged at the point $O$ and held in place vertically between two walls using two massless springs of same spring constant $K$. The springs are connected at the midpoint and at the top-end $(O^{\prime})$ of the rod,as shown in Fig. $1$,and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $f_1$. On the other hand,if both the springs are connected at the midpoint of the rod,as shown in Fig. $2$,and the rod is made to oscillate by a small angular displacement,then the frequency of oscillation is $f_2$. Ignoring gravity and assuming motion only in the plane of the diagram,the value of $\frac{f_1}{f_2}$ is:

$A$ body performs linear $S$.$H$.$M$. with amplitude $a$. When it is at a distance $\frac{a}{3}$ from the extreme position,the magnitude of velocity is $\frac{1}{3}$ times the magnitude of acceleration. The period of $S$.$H$.$M$. is:

The displacement-time graph of a particle executing $SHM$ is shown. Which of the following statements is wrong?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo