1403 Energy In SHM

The total mechanical energy of the system is conserved.

$\begin{aligned} E = \frac{1}{2}mv_x^2 + \frac{1}{2}kx^2 = \frac{1}{2} kA^2 \quad \text{(1)} \end{aligned}$

$E$ - Total mechanical energy in SHM.
$v$ - Velocity.
$k$ - Force constant of restoring force.
$A$ - Amplitude.
$x$ - Displacement.
From Eq. (1), we can solve it for $v_x$

$\begin{aligned} v_x &= \pm \sqrt{\frac{k}{m}} \sqrt{A^2 - x^2}\\ v_{max} &= \sqrt{\frac{k}{m}} A = \omega A \end{aligned}$

Exercises

25, 27, 34

(14.25) A small block is attached to an ideal spring and is mov ing in SHM on a horizontal frictionless surface. The amplitude of the motion is 0.165 m. The maximum speed of the block is $3.90$ m/s. What is the maximum magnitude of the acceleration of the block?

Solution

$\begin{aligned} v_{max} &= \omega A \\ \To \omega &= \frac{v_{max}}{A} = \frac{3.90}{0.165} rad/s\\ a_{max} &= \omega^2 A = 92.2 rad/s^2 \end{aligned}$

(14.27) A 0.150-kg toy is undergoing SHM on the end of a horizontal spring with force constant k = 300 N/m. When the toy is 0.0120 m from its equilibrium position, it is observed to have a speed of 0.400 m/s. What are the toy's (a) total energy at any point of its motion; (b) amplitude of motion; (c) maximum speed duringits motion?

Solution

Todo
a. $E = 0.0336J$
b. $A = 0.0150m$
c. $v_{max} = 0.669 m/s$

(14.34) A 2.00-kg frictionless block is attached to an ideal spring with force constant 315 N/m. Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude of the motion, (b) the block's maximum acceleration, and (c) the maximum force the spring exerts on the block.

Solution

Todo