A bob of a simple pendulum of mass m performs | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The kinetic energy $(K.E.)$ of a particle performing $SHM$ is given by: $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$ Given $x = \frac{A}{2}$,substi

Vedclass · Accepted Answer

$\frac{3 m \pi^2 A^2}{2 T^2}$

Vedclass · Answer

(D) The kinetic energy $(K.E.)$ of a particle performing $SHM$ is given by: $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$ Given $x = \frac{A}{2}$,substitute this into the equation: $K.E. = \frac{1}{2} m \omega^2 (A^2 - (\frac{A}{2})^2)$ $K.E. = \frac{1}{2} m \omega^2 (A^2 - \frac{A^2}{4}) = \frac{1}{2} m \omega^2 (\frac{3 A^2}{4}) = \frac{3}{8} m \omega^2 A^2$ Since the angular frequency $\omega = \frac{2 \pi}{T}$,substitute this value: $K.E. = \frac{3}{8} m (\frac{2 \pi}{T})^2 A^2$ $K.E. = \frac{3}{8} m (\frac{4 \pi^2}{T^2}) A^2$ $K.E. = \frac{3 m \pi^2 A^2}{2 T^2}$

$A$ bob of a simple pendulum of mass $m$ performs $SHM$ with amplitude $A$ and period $T$. The kinetic energy of the pendulum at displacement $x = \frac{A}{2}$ will be:

Similar Questions

The total mechanical energy of a particle in $SHM$ is

The average energy possessed by an oscillator at a temperature $300 \ K$ is (Boltzmann constant $= 1.38 \times 10^{-23} \ J K^{-1}$).

For a particle $P$ performing $S.H.M.$, when displacement is $x$, potential energy and restoring force acting on it are denoted by $E$ and $F$ respectively. The relation between $x, E$, and $F$ is:

$A$ mass of $1 \text{ kg}$ is executing $SHM$. Its displacement is given by $x = 6.0 \cos(100t + \pi/4) \text{ cm}$. What is the maximum kinetic energy (in $text{ J}$)?

For a body performing simple harmonic motion,its potential energy is $E_x$ at displacement $x$ and $E_y$ at displacement $y$ from the mean position. The potential energy $E_0$ at displacement $(x+y)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module