Simple harmonic oscillations — practice questions

**Step 1: $Calculate angular frequency ($\omega$).**$ The formula is $\omega = 2\pi / T$. $\omega = 2\pi / 1.2 \text{ s} \approx 5.236 \text{ rad/s}$. Step 2: Calculate maximum velocity ($v_0$). The formula is $v_0 = \omega x_0$. Remember to convert amplitude to metres! $x_0 = 5.0 \text{ cm} = 0.050 \text{ m}$. $v_0 = 5.236 \text{ rad/s} \times 0.050 \text{ m} \approx 0.2618 \text{ m/s}$. Step 3: Calculate acceleration (a) at $x = 3.0 \text{ cm}$. The formula is $a = -\omega^2 x$. Convert displacement to metres. $x = 3.0 \text{ cm} = 0.030 \text{ m}$. $a = -(5.236 \text{ rad/s})^2 \times 0.030 \text{ m} \approx -0.822 \text{ m/s}^2$.

**Step 1: $Calculate angular frequency ($\omega$).**$ The relationship between period T and angular frequency $\omega is$\omega = 2\pi / T$$. $\omega = 2\pi / 0.80 \text{ s} = 7.854 \text{ rad/s}$.

Step 2: Calculate the total energy of the system ($E_{total}$). The total energy in SHM is constant and can be calculated at the point of maximum displacement (amplitude), where all energy is potential. The formula is $E_{total} = \frac{1}{2} m \omega^2 x_0^2$. Remember to convert amplitude to metres: $x_0 = 10 \text{ cm} = 0.10 \text{ m}$. $E_{total} = \frac{1}{2} \times 0.50 \text{ kg} \times (7.854 \text{ rad/s})^2 \times (0.10 \text{ m})^2$ $E_{total} = 0.25 \times 61.685 \times 0.01 \approx 0.1542 \text{ J}$. Rounding to two significant figures, $E_{total} = 0.15 \text{ J}$.

Step 3: Calculate the kinetic energy (KE) at $x = 6.0 \text{ cm}$. First, convert displacement to metres: $x = 6.0 \text{ cm} = 0.060 \text{ m}$. We know that total energy is the sum of kinetic and potential energy: $E_{total} = KE + PE$. Therefore, $KE = E_{total} - PE$. The potential energy at displacement x is $PE = \frac{1}{2} m \omega^2 x^2$. $PE = \frac{1}{2} \times 0.50 \text{ kg} \times (7.854 \text{ rad/s})^2 \times (0.060 \text{ m})^2$ $PE = 0.25 \times 61.685 \times 0.0036 \approx 0.0555 \text{ J}$. Now, calculate KE: $KE = 0.1542 \text{ J} - 0.0555 \text{ J} \approx 0.0987 \text{ J}$. Rounding to two significant figures, $KE = 0.099 \text{ J}$.