Q.In Exercise 13.9, let us take the position of mass when the spring is unstretched as $x = 0$, and the direction from left to right as the positive direction of $x$-axis. Give $x$ as a function of time $t$ for the oscillating mass if at the moment we start the stopwatch ($t = 0$), the mass is

Concept: Simple Harmonic Motion — the solution to $m\frac{d^2x}{dt^2} = -kx$ is $x(t) = A\sin(\omega t + \phi)$ or $x(t) = A\cos(\omega t + \phi)$, where $\omega = \sqrt{k/m}$.

Reasoning:

1. For SHM, the general form is $x(t) = A\sin(\omega t + \phi)$ or $x(t) = A\cos(\omega t + \phi)$. Amplitude $A$ and angular frequency $\omega$ are fixed by the system (spring constant $k$, mass $m$). Only the initial phase $\phi$ changes with the starting condition.
2. (a) At $t=0$, mass at mean position ($x=0$) and moving to the right (positive velocity). Using $x = A\sin(\omega t)$ gives $x(0)=0$ and $v(0)=A\omega\cos(0)=A\omega>0$. So $x(t) = A\sin\omega t$.
3. (b) At $t=0$, mass at maximum stretched position ($x = +A$) and momentarily at rest. Using $x = A\cos\omega t$ gives $x(0)=A$, $v(0)=0$. So $x(t) = A\cos\omega t$. …

The Core Idea: Why Phase Alone Changes

Simple Harmonic Motion is the projection of uniform circular motion onto a diameter. The general solution is $x(t) = A\sin(\omega t + \phi)$ or equivalently $x(t) = A\cos(\omega t + \phi')$ — the two forms differ by a constant phase shift of $\pi/2$. What matters physically is that the amplitude $A$ (maximum displacement) and angular frequency $\omega$ (determined by the spring constant $k$ and mass $m$, $\omega = \sqrt{k/m}$) are fixed by the system, not by how we start it. The initial conditions — where the mass is and how it's moving at $t=0$ — only fix the constant $\phi$.

So all three parts of this question share the same $A$ and $\omega$. The only difference is the starting point on the oscillation cycle, which is captured by $\phi$.

Step-by-Step Solution

We take $x=0$ as the unstretched (mean) position, positive $x$ to the right. Let the amplitude be $A$ and angular frequency $\omega = \sqrt{k/m}$.

1. Case (a): Mass at the mean position at $t=0$

At $t=0$, $x=0$. The mass is passing through equilibrium. In SHM, when $x=0$, the velocity is maximum. We need a function that gives $x=0$ at $t=0$.

The sine function works naturally: $\sin 0 = 0$. So we write:

Here the initial phase $\phi = 0$ (if using the sine form). The velocity $v = A\omega\cos(\omega t)$ is maximum positive at $t=0$, meaning the mass is moving to the right through the mean position.

2. Case (b): Mass at the maximum stretched position at $t=0$

"Maximum stretched" means the spring is pulled to the right as far as it goes — so $x = +A$ at $t=0$. We need a function that gives $x = A$ when $t=0$.

The cosine function works: $\cos 0 = 1$. So:

Here the initial phase $\phi = 0$ (in the cosine form). At $t=0$, the velocity $v = -A\omega\sin(\omega t)$ is zero — the mass is momentarily at rest at the extreme right, about to move left.

3. Case (c): Mass at the maximum compressed position at $t=0$

"Maximum compressed" means the spring is pushed to the left as far as it goes — so $x = -A$ at $t=0$. We need $x = -A$ when $t=0$. …

Step 1: In every case, the SHM has the same $\omega=\sqrt{k/m}$ and same amplitude $A$ (fixed by the spring, mass, and the 2.0 cm pull in Exercise 13.9) — only the starting point on the cycle differs, so only the phase changes.

Step 2 — (a) mean position at $t=0$: $x(0)=0$ with the mass moving in the $+x$ direction matches the sine function directly: $x(t)=A\sin\omega t$ (since $\sin0=0$ and its derivative is maximum there).

Step 3 — (b) maximum stretched at $t=0$: $x(0)=+A$, momentarily at rest, matches the cosine function: $x(t)=A\cos\omega t$ (since $\cos0=1$ and its derivative is zero there). …