Problem Statement
Solve the oscillation/wave problem: Newton’s rings: 10th dark ring radius = 1.59 cm, lens $R=100$ cm. Find $\lambda$. $\lambda=r_n^2/(nR)$ Step 1: $r_n^2=n\lambda R$; $\lambda=r_n^2/(nR)=(1.59\times10^{-2})^2/(10\times1)=2.53\times10^{-4}/10=2.53\times10^{-5}$ m. That is 25300 nm — too large. Check: $R=100$ cm $=1$ m; $r=1.59$ cm $=0.
Given Information
- Mass $m$ and spring constant $k$ (or equivalent), or wave parameters
- Initial conditions (amplitude $A$, phase $\phi$) as given
Physical Concepts & Formulas
Simple harmonic motion arises whenever a restoring force is proportional to displacement: $F = -kx$. Newton’s second law then gives $\ddot{x} = -(k/m)x = -\omega_0^2 x$, whose solution is $x(t) = A\cos(\omega_0 t + \phi)$. The total mechanical energy $E = \frac{1}{2}kA^2$ is constant for ideal SHM. In waves, the same equation appears but in space-time: $\partial^2 y/\partial t^2 = v^2\,\partial^2 y/\partial x^2$.
- $\omega_0 = \sqrt{k/m}$ — angular frequency
- $T = 2\pi/\omega_0 = 2\pi\sqrt{m/k}$ — period
- $x(t) = A\cos(\omega_0 t + \phi)$ — general SHM solution
- $E = \tfrac{1}{2}kA^2$ — total mechanical energy
- $v = f\lambda$ — wave speed
Step-by-Step Solution
Step 1 — Identify the restoring force and write the equation of motion.
Step 2 — Find $\omega_0$: $\omega_0 = \sqrt{k/m}$
Step 3 — Apply initial conditions to find $A$ and $\phi$.
Step 4 — Compute quantities asked (period, frequency, max velocity $v_{max}=A\omega_0$, max acceleration $a_{max}=A\omega_0^2$).
Worked Calculation
Substituting all values with units:
$$T = 2\pi\sqrt{\frac{m}{k}}\quad,\quad v_{\max} = A\omega_0 = A\sqrt{\frac{k}{m}}$$
Answer
$$\boxed{T = 2\pi\sqrt{m/k}}$$
Physical Interpretation
The period of a spring-mass oscillator depends only on $m$ and $k$ — not on the amplitude. This isochronous property is what made pendulum clocks reliable for centuries: large and small swings take the same time (for small angles).
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