we have a pendulum with an object at the end with a mass of 0.120kg and the length of the pendulum is 0.44m, this pendulum oscillates with an amplitude of 4°,What is the total energy (kinetic plus potential) of this pendulum
lest find the kinetic energy first:
$$KE = \frac{1}{2}mv^2$$
in order to find the velocity first we find the period:
$$T = 2\pi \sqrt{\frac{l}{g}} = 1.331 $$
angular frequency is:
$$\omega = \frac{2\pi}{T} = 4.72 $$
next, we find the velocity I convert the 4° angular frequency times amplitude is velocity
$$v = A \omega = (0.069)(4.72)=0.325$$
$$KE = \frac{1}{2}mv^2 = 6.33\times 10^{-3}$$
now let's find the gravitational potential energy:
$$PE = mgL(1-cos\theta) = 1.26\times 10^{-3}$$
$$E= KE +PE= 7.59\times 10^{-3}$$
this answer is wrong book answer is 1.3 X 10^-3 J
What did I get wrong also if you have a more efficient way of doing the problem I would like to see it
Best Answer
First, you're overcomplicating things. The total energy of the pendulum will be equal at all points of its swing, so the simplest way to find it is to calculate the PE at the ends of its swing, when the KE is equal to zero.
You already did this:
$$PE = mgL(1-\cos\theta) \approx 1.26 \times 10^{-3} \text{J}$$
which is the answer to the question.
You cannot calculate the KE at the bottom of the swing and then add it to the PE at the top of the swing as you're doing because the two energies are being calculated at different points in the swing.
You could just calculate the KE at the bottom of the swing (when the PE will be zero) to get the same result, but in your calculations for KE you're making a couple of mistakes:
To calculate the KE at the bottom of the swing, start with the simple harmonic motion equation
$$\theta (t) = \theta_0 \cos \left( \frac{2\pi}{T} t \right)$$
Deriving this gives
$$\omega (t) = \frac{-2\pi\theta_0}{T} \sin \left( \frac{2\pi}{T} t \right)$$
Now using the fact that at the bottom of the first swing, $t = \frac{T}{4}$, we get
$$\begin{align} \omega_{\text{max}} (t) &= \rule[-25px]{1px}{55px} \frac{-2\pi\theta_0}{T} \sin \left[ \frac{2\pi}{T} \left( \frac{T}{4} \right) \right] \;\rule[-25px]{1px}{55px} \\[2mm] &= \rule[-20px]{1px}{50px} \frac{-2\pi\theta_0}{T} \sin \left( \frac{\pi}{2} \right) \; \rule[-20px]{1px}{50px} \\[2mm] &= \frac{2\pi\theta_0}{T} \end{align}$$
and finally
$$\begin{align} v_{\text{max}} &= L \omega_{\text{max}} \\[2mm] &= \frac{2\pi\theta_0 L}{T} \end{align}$$
Using $\theta_0 = 4^{\circ} = 0.0698$ rads and your calculated value $T=1.331$ yields
$$\begin{align} v_{\text{max}} &= \frac{2\pi (0.0698)(0.44)}{1.331} \\[2mm] &\approx 0.145 \text{m/sec} \end{align}$$
and so finally the KE at the bottom of the swing will be
$$\begin{align} KE &= \frac{1}{2}mv^2 \\[2mm] &\approx 1.26 \times 10^{-3} \text{J} \end{align}$$
which agrees with the previously calculated PE value.