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# Part (a) – Use the formula for the period of a spring.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] T = 2\pi\sqrt{\frac{m}{k}} [/katex] | The formula for the period [katex] T [/katex] of oscillation of a mass-spring system, where [katex] m [/katex] is the mass and [katex] k [/katex] is the spring constant. |
| 2 | Substitute [katex] m = 2 \, \text{kg} [/katex], [katex] k = 100 \, \text{N/m} [/katex] | Insert the values into the formula to calculate the period. |
| 3 | [katex] T = 2\pi\sqrt{\frac{2}{100}} = 2\pi\sqrt{0.02} \approx 0.89 \, \text{s} [/katex] | Calculating the square root and then multiplying by [katex] 2\pi [/katex] gives the period. This calculation is independent of the amplitude of the oscillation (0.5 m or 1 m), assuming no damping and small angle. |
| 4 | [katex] T \approx 0.89 \, \text{s} [/katex] | This is the period of oscillation for any initial stretching (i) 0.5 m and (ii) 1 m, since [katex] T [/katex] does not depend on amplitude for simple harmonic motion in ideal springs. |
# Part (b) – The relationship between mass and period.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] T’ = 2\pi\sqrt{\frac{m’}{k}} [/katex]. | If the mass [katex] m [/katex] is doubled, we substitute [katex] m’ = 2m = 4 \, \text{kg} [/katex] into the formula for period. |
| 2 | Substitute [katex] m’ = 4 \, \text{kg} [/katex], [katex] k = 100 \, \text{N/m} [/katex]. | Insert the new mass into the period formula. |
| 3 | [katex] T’ = 2\pi\sqrt{\frac{4}{100}} = 2\pi\sqrt{0.04} \approx 1.26 \, \text{s} [/katex]. | Calculating the square root and multiplying by [katex] 2\pi [/katex] gives the new period. |
| 4 | [katex] T’ \approx 1.26 \, \text{s} [/katex] | This is the new period of oscillation when the mass is doubled. The period increases, showing that it is dependent on mass. |
# Part (c) – Justification.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] F = -kx [/katex] | The restoring force [katex] F [/katex] provided by the spring is proportional to the displacement [katex] x [/katex] and the spring constant [katex] k [/katex], according to Hooke’s Law. |
| 2 | Increasing mass requires a longer time to accelerate/decelerate due to spring force. | For the same displacement, the spring force [katex] F [/katex] remains constant. When mass is doubled, although the same force acts on a larger mass, resulting in a lower acceleration ([katex] a = \frac{F}{m} [/katex]), leading to a longer period of oscillation. |
| 3 | Summary: Greater mass leads to reduced acceleration from the same force, increasing time for one complete oscillation. | This explains why doubling the mass increases the period as shown in the earlier calculation. It showcases the inverse relationship between acceleration and mass in the context of Newton’s second law. |
Just ask: "Help me solve this problem."
A 20 g piece of clay moving at a speed of 50 m/s strikes a 500 g pendulum bob at rest. The length of a string is 0.8 m. After the collision the clay-bob system starts to oscillate as a simple pendulum.
What is the effect on the period of a pendulum if you double its length?
A small block moving with a constant speed v collides inelastically with a block M attached to one end of a spring k. The other end of the spring is connected to a stationary wall. Ignore friction between the blocks and the surface.
A pendulum has a period of 2.0 s on Earth. What is its length?
When the mass of a simple pendulum is tripled, the time required for one complete vibration
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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