Problem Statement
Solve the oscillation/wave problem: X-rays have frequencies ranging from $3\times10^{16}$ Hz to $3\times10^{19}$ Hz. Find the corresponding range of wavelengths. $\lambda = c/f$ Step 1: For $f_{min} = 3\times10^{16}$ Hz (longest wavelength): $$\lambda_{max} = \frac{3\times10^8}{3\times10^{16}} = 10^{-8}\text{ m} = 10\text{ nm}$$ Step
Given Information
- $\lambda_{max} = \frac{3\times10^8}{3\times10^{16}} = 10^{-8}\text{ m} = 10\text{ nm}$
Physical Concepts & Formulas
Projectile motion decomposes into independent horizontal and vertical components. Horizontal: constant velocity (no air resistance). Vertical: constant downward acceleration $g$. The trajectory is a parabola. Maximum range occurs at $45°$ launch angle; max height at $90°$.
- $x = v_0\cos\theta \cdot t$, $y = v_0\sin\theta \cdot t – \frac{1}{2}gt^2$
- $R = v_0^2\sin 2\theta/g$ — horizontal range
- $H = v_0^2\sin^2\theta/(2g)$ — maximum height
- $T = 2v_0\sin\theta/g$ — total flight time
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\lambda_{max} = \frac{3\times10^8}{3\times10^{16}} = 10^{-8}\text{ m} = 10\text{ nm}$$
$$T = 2\pi\sqrt{\frac{m}{k}}\quad,\quad v_{\max} = A\omega_0 = A\sqrt{\frac{k}{m}}$$
$$\boxed{T = 2\pi\sqrt{m/k}}$$
Answer
$$\boxed{T = 2\pi\sqrt{m/k}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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