(Consider mass of the bob = $20 \text{ g}$)
The total mechanical energy ($E$) of a simple pendulum is the sum of its kinetic energy ($KE$) and potential energy ($PE$): $E = KE + PE$
We are given that the total energy is $E = 0.02 \text{ J}$.
At the equilibrium position, the potential energy ($PE$) is minimum, typically considered zero. Therefore, at this point, all the energy is kinetic energy ($KE_{max}$). $E = KE_{max} + 0$ $KE_{max} = E = 0.02 \text{ J}$
The formula for kinetic energy is $KE = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the speed.
Given:
At equilibrium, $KE_{max} = \frac{1}{2}mv^2$. We set this equal to the total energy:
$0.02 \text{ J} = \frac{1}{2} \times (0.02 \text{ kg}) \times v^2$
Now, solve for $v$:
$0.02 = 0.01 \times v^2$
$v^2 = \frac{0.02}{0.01}$
$v^2 = 2$
$v = \sqrt{2} \text{ m/s}$
Calculating the square root:
$v \approx 1.414 \text{ m/s}$
Rounding to two decimal places, the speed is approximately $1.41 \text{ m/s}$.
For a travelling harmonic wave $y(x, t) = 2.0 \cos 2\pi(10 t - 0.0080 x + 0.35)$, where $x$ and $y$ are in cm and $t$ in s. The phase difference between oscillatory motion of two points separated by a distance of $0.5 \text{ m}$ is :