Periodic Motion: A Calculus Approach

P1. SHM Kinematics

This problem applies the definitions of $\omega$, $\phi$, and uses the energy/velocity relationship to find the position.

Part (a): Angular Frequency $\omega$

The angular frequency is determined by the system's inertia ($m$) and stiffness ($k$) as $\omega = \sqrt{k/m}$:

$$ \omega = \sqrt{\frac{50 \text{ N/m}}{0.5 \text{ kg}}} = \sqrt{100} = 10 \text{ rad/s} $$

Part (b): Phase Constant $\phi$

We use the position equation (3) with the initial condition $x(0) = A$:

$$ x(0) = A \cos(\omega(0) + \phi) \implies A = A \cos(\phi) $$

For this to be true, $\cos(\phi)$ must equal 1. The simplest solution is:

$$ \phi = 0 $$

Part (c): Position $x$ when $v = v_{max}/2$

The velocity equation derived from energy conservation is $v = \pm \omega \sqrt{A^2 - x^2}$. We know $v_{max} = A\omega$. We set $v = \frac{1}{2} A\omega$ and solve for $x$:

$$ \begin{aligned} \frac{1}{2} A\omega &= \omega \sqrt{A^2 - x^2} \\ \frac{A}{2} &= \sqrt{A^2 - x^2} \\ \frac{A^2}{4} &= A^2 - x^2 \\ x^2 &= A^2 - \frac{A^2}{4} = \frac{3}{4} A^2 \\ x &= \pm \frac{\sqrt{3}}{2} A \\ x &= \pm \frac{\sqrt{3}}{2} (0.1 \text{ m}) \approx \pm 0.0866 \text{ m} \end{aligned} $$

The speed is half of the maximum when the mass is $86.6\%$ of the amplitude away from equilibrium.

P2. SHM Time Period

This problem uses the explicit kinematic solution (3) to find the angular frequency, which then yields the period and the spring constant.

Part (a): Time Period $T$

By comparing the given equation $x(t) = 0.05 \cos(4t)$ to the standard form $x(t) = A \cos(\omega t + \phi)$, we identify the angular frequency:

$$ \omega = 4 \text{ rad/s} $$

The period $T$ is related to $\omega$ by $T = 2\pi/\omega$:

$$ T = \frac{2\pi}{4 \text{ rad/s}} = \frac{\pi}{2} \text{ s} \approx 1.57 \text{ s} $$

Part (b): Spring Constant $k$

The angular frequency is also defined as $\omega = \sqrt{k/m}$. Solving for $k$ gives $k = m\omega^2$.

$$ k = (2 \text{ kg}) (4 \text{ rad/s})^2 = (2) (16) = 32 \text{ N/m} $$

P3. Energy to Amplitude

This uses the two key formulas for total mechanical energy in SHM: $E$ at max displacement ($U_{max}$) and $E$ at equilibrium ($K_{max}$).

Part (a): Amplitude $A$

The total energy $E$ is equal to the maximum potential energy, $E = U_{max} = \frac{1}{2} k A^2$ (8). We solve for $A$:

$$ A = \sqrt{\frac{2E}{k}} = \sqrt{\frac{2(12 \text{ J})}{300 \text{ N/m}}} = \sqrt{\frac{24}{300}} = \sqrt{0.08} \approx 0.283 \text{ m} $$

Part (b): Maximum Velocity $v_{max}$

The total energy $E$ is also equal to the maximum kinetic energy, $E = K_{max} = \frac{1}{2} m v_{max}^2$. We solve for $v_{max}$:

$$ v_{max} = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2(12 \text{ J})}{1.5 \text{ kg}}} = \sqrt{\frac{24}{1.5}} = \sqrt{16} $$ $$ v_{max} = 4.0 \text{ m/s} $$

P4. Average Kinetic Energy

This proof requires integration over one full period $T$ to find the average value $\langle K \rangle$ and relies on the $\sin^2\theta$ integral identity.

Step 1: Define the Average Value

The average value of any time-dependent quantity $f(t)$ over a period $T$ is $\langle f \rangle = \frac{1}{T} \int_{0}^{T} f(t) dt$. Using the kinetic energy function (7) with $k=m\omega^2$ and factoring out constant terms:

$$ \langle K \rangle = \frac{1}{T} \int_{0}^{T} \left( \frac{1}{2} k A^2 \sin^2(\omega t) \right) dt = \frac{1}{2} k A^2 \frac{1}{T} \int_{0}^{T} \sin^2(\omega t) dt $$

Step 2: Apply Trigonometric Identity and Integrate

We use the identity $\sin^2\theta = \frac{1}{2} (1 - \cos(2\theta))$. Over a full period $T$, the integral of the cosine term $\cos(2\omega t)$ is zero.

$$ \begin{aligned} \langle K \rangle &= \frac{1}{4} k A^2 \frac{1}{T} \int_{0}^{T} (1 - \cos(2\omega t)) dt \\ \langle K \rangle &= \frac{1}{4} k A^2 \frac{1}{T} \int_{0}^{T} 1 dt = \frac{1}{4} k A^2 \frac{T}{T} \\ \langle K \rangle &= \frac{1}{4} k A^2 \end{aligned} $$

Conclusion

Since the total energy is $E = \frac{1}{2} k A^2$ (8), we conclude that $\langle K \rangle = \frac{1}{2} E$.

P5. Position-Dependent Force

This requires applying Newton's Second Law and comparing the resulting differential equation to the standard form of the SHM equation (2).

Step 1: Form the Differential Equation (ODE)

Starting with Newton's Second Law ($F = m a$) and substituting the given force $F = -4\beta x$ and $a = \frac{d^2x}{dt^2}$:

$$ m \frac{d^2x}{dt^2} = -4\beta x $$

Rearrange into the standard SHM ODE form $\frac{d^2x}{dt^2} + \omega^2 x = 0$:

$$ \frac{d^2x}{dt^2} + \left( \frac{4\beta}{m} \right) x = 0 $$

Since the acceleration is proportional to the negative of the displacement (which requires $\beta$ to be positive), the motion is Simple Harmonic Motion (SHM).

Step 2: Find Angular Frequency $\omega$ and Period $T$

By comparison with Equation (2), the angular frequency squared is:

$$ \omega^2 = \frac{4\beta}{m} \implies \omega = \sqrt{\frac{4\beta}{m}} = 2 \sqrt{\frac{\beta}{m}} $$

The period $T$ is $T = 2\pi/\omega$:

$$ T = \frac{2\pi}{2 \sqrt{\frac{\beta}{m}}} = \pi \sqrt{\frac{m}{\beta}} $$

P6. Logarithmic Decrement

This problem analyzes the exponential decay factor of the underdamped solution (10) over exactly one damped period $T_d = 2\pi/\omega_d$.

Step 1: Define Successive Amplitudes

For underdamped motion, the amplitude $A(t)$ is $A_0 e^{-\beta t}$. Let $A_n$ be the amplitude at time $t_n$. The next amplitude, $A_{n+1}$, occurs exactly one period later, at $t_{n+1} = t_n + T_d$.

$$ \begin{aligned} A_n &= A_0 e^{-\beta t_n} \\ A_{n+1} &= A_0 e^{-\beta (t_n + T_d)} = A_0 e^{-\beta t_n} e^{-\beta T_d} \end{aligned} $$

Step 2: Form the Amplitude Ratio

The ratio of successive maximum amplitudes is formed by dividing $A_n$ by $A_{n+1}$. The common terms cancel:

$$ \frac{A_n}{A_{n+1}} = \frac{A_0 e^{-\beta t_n}}{A_0 e^{-\beta t_n} e^{-\beta T_d}} = e^{\beta T_d} $$

Step 3: Calculate the Logarithmic Decrement $\delta$

The logarithmic decrement $\delta$ is the natural logarithm of this ratio:

$$ \delta = \ln \left( \frac{A_n}{A_{n+1}} \right) = \ln \left( e^{\beta T_d} \right) $$

Using the property $\ln(e^x) = x$, we derive the required result:

$$ \delta = \beta T_d $$

P7. Critically Damped Max Displacement

For critically damped motion (11), the maximum displacement $t_{max}$ occurs when the velocity $v(t) = \frac{dx}{dt}$ is zero. This requires solving for the constants $A$ and $B$ using the initial conditions.

Step 1: Solve for Constants $A$ and $B$ using Initial Conditions

The solution form is $x(t) = (A + B t) e^{-\omega_0 t}$.

Initial position $x(0) = 0$:

$$ x(0) = (A + 0) e^0 = A \implies A = 0 $$

Thus, $x(t) = B t e^{-\omega_0 t}$. Now, we find the velocity $v(t)$ by differentiation (using the product rule):

$$ \begin{aligned} v(t) &= \frac{dx}{dt} = B \left[ (1) e^{-\omega_0 t} + t (-\omega_0 e^{-\omega_0 t}) \right] \\ v(t) &= B e^{-\omega_0 t} (1 - \omega_0 t) \end{aligned} $$

Initial velocity $v(0) = v_0$:

$$ v(0) = B e^0 (1 - 0) = B \implies B = v_0 $$

The specific displacement equation is $x(t) = v_0 t e^{-\omega_0 t}$.

Step 2: Find Time $t_{max}$ when $\frac{dx}{dt} = 0$

Set the velocity function to zero to find the time of maximum displacement, $t_{max}$:

$$ v_0 e^{-\omega_0 t_{max}} (1 - \omega_0 t_{max}) = 0 $$

Since $v_0 \ne 0$ and the exponential term $e^{-\omega_0 t}$ is never zero, we must have:

$$ 1 - \omega_0 t_{max} = 0 $$ $$ t_{max} = \frac{1}{\omega_0} $$

P8. Energy in Damped Motion

For underdamped motion, the energy decays due to the damping term. We show the proportionality by considering the energy at the maximum displacement, where $E \approx U_{max}$.

Step 1: Define the Instantaneous Amplitude $A(t)$

The general solution (10) can be seen as an oscillating part $\cos(\omega_d t + \phi)$ multiplied by a time-dependent amplitude $A(t)$:

$$ A(t) = A_0 e^{-\beta t} $$

Step 2: Relate Instantaneous Energy $E(t)$ to Amplitude $A(t)$

For SHM, the total energy $E$ is proportional to the square of the amplitude, $E = \frac{1}{2} k A^2$ (8). For slowly decaying, underdamped motion, the instantaneous energy $E(t)$ at any time is proportional to the square of the instantaneous amplitude $A(t)$:

$$ E(t) \propto A(t)^2 $$

Step 3: Show the Proportionality

Substitute the expression for $A(t)$ into the proportionality relationship:

$$ E(t) \propto \left( A_0 e^{-\beta t} \right)^2 $$ $$ E(t) \propto A_0^2 e^{-2\beta t} $$

Since $A_0^2$ is a constant, the time-dependence of the energy is entirely determined by the exponential decay factor. Thus, the instantaneous energy decays proportionally to $e^{-2\beta t}$.

P9. Two-Block System

This is a dynamics problem that requires finding the maximum acceleration of the two-block system and relating it to the maximum static friction force.

Step 1: System Angular Frequency $\omega$

Since the blocks move together, the total oscillating mass is $(M+m)$. The angular frequency of the system is:

$$ \omega^2 = \frac{k}{M+m} $$

Step 2: Maximum Acceleration $a_{max}$ of the System

The magnitude of the maximum acceleration of the system is $a_{max} = A_{max} \omega^2$ (from (5)).

$$ a_{max} = A_{max} \left( \frac{k}{M+m} \right) $$

Step 3: Force Required by Top Block $m$

For the top block $m$ to move with this acceleration, it requires a net force $F_{net, m}$ provided by static friction $F_f$.

$$ F_{net, m} = m a_{max} $$

Step 4: Maximum Friction Limit

To prevent slipping, the required force $F_{net, m}$ must be less than or equal to the maximum static friction $F_{f, \text{max}} = \mu_s F_N$. Since $F_N = mg$:

$$ F_{net, m} \le \mu_s m g \implies m A_{max} \omega^2 \le \mu_s m g $$

Step 5: Solve for $A_{max}$

Cancel the mass $m$ and substitute $\omega^2$ from Step 1:

$$ A_{max} \left( \frac{k}{M+m} \right) = \mu_s g $$ $$ A_{max} = \frac{\mu_s g (M+m)}{k} $$

P10. Time-Dependent Potential

The motion near an equilibrium point of a complex potential can be approximated as SHM by performing a Taylor expansion of the potential energy $U(x)$ around that point.

Step 1: Find Equilibrium Point $x_0$

Equilibrium occurs where the force $F(x) = -\frac{dU}{dx} = 0$.

$$ \frac{dU}{dx} = k x + 3\lambda x^2 $$

Setting this to zero gives equilibrium points $x_0 = 0$ and $x_0 = -k/(3\lambda)$. We analyze the motion near $\mathbf{x_0 = 0}$.

Step 2: Calculate the Effective Spring Constant $k_{eff}$

For small oscillations, the potential $U(x)$ is approximated by $U(x) \approx \frac{1}{2} k_{eff} (x-x_0)^2$, where $k_{eff}$ is the second derivative of $U$ evaluated at the equilibrium point:

$$ k_{eff} = \left. \frac{d^2U}{dx^2} \right|_{x=x_0} $$ $$ \frac{d^2U}{dx^2} = k + 6\lambda x $$

Evaluating at $x_0 = 0$:

$$ k_{eff} = k + 6\lambda(0) = k $$

Step 3: Determine the Effective Angular Frequency $\omega_{eff}$

The effective angular frequency is calculated using the effective spring constant $k_{eff}$ in the standard SHM frequency formula $\omega = \sqrt{k/m}$:

$$ \omega_{eff} = \sqrt{\frac{k_{eff}}{m}} = \sqrt{\frac{k}{m}} $$

The nonlinear cubic term $\lambda x^3$ only contributes to the third-order derivative, meaning it does not affect the effective angular frequency for infinitesimally small oscillations near $x=0$.