Spring oscillations: build the motion from the initial state

1. Read the model and the question

A block of mass $m=0.50\,\mathrm{kg}$ is attached to an ideal horizontal spring of stiffness $k=2.0\,\mathrm{N/m}$, with the other end fixed. The spring obeys Hooke’s law throughout the motion considered. Ignore friction, damping and driving. Measure displacement $x$ from equilibrium, positive to the right. Velocity $v$ is signed: positive means moving right. Acceleration is $a$, and time $t$ is measured from the specified initial instant.

At $t=0$,

Find the motion equation and the first time after $t=0$ that the block crosses equilibrium. State its velocity at that crossing.

The vertical forces balance. Horizontally, the restoring spring force is $-kx$, so Newton’s second law gives

Here $\omega$ is the angular frequency of the oscillation; $T=2\pi/\omega$ is the period, the time for one complete cycle. Initially $a_0=-0.120\,\mathrm{m/s^2}$. The block is moving right but slowing down. It must reach a right-hand turning point before returning to equilibrium. Constant-acceleration formulas do not apply: the acceleration changes with position.

2. Use both initial values

For this free simple harmonic motion, the position and velocity can be written

The constants $C,D$ have units of length. At $t=0$, cosine is 1 and sine is 0. Therefore $C=x_0$ and $D=v_0/\omega$, giving $C=0.030\,\mathrm m$, $D=0.040\,\mathrm m$.

With numerical time $t$ in seconds,

Substitute $t=0$ to check both initial values. Using only $x=x_0\cos(\omega t)$ would give zero initial velocity, contrary to this question.

3. Recover amplitude and phase

We can also write $x=A\cos(\omega t+\phi)$. The amplitude $A\ge0$ is the greatest displacement magnitude from equilibrium; $\phi$ is the phase at the chosen initial instant. Use the cosine addition formula:

Compare the coefficients with $x=C\cos(\omega t)+D\sin(\omega t)$: $C=A\cos\phi$ and $D=-A\sin\phi$. The minus sign comes from the cosine addition formula. Since $\cos^2\phi+\sin^2\phi=1$, we have $C^2+D^2=A^2$, so

Here $\cos\phi>0$ and $\sin\phi<0$, so choose a phase in the fourth quadrant. The reference angle is $\alpha=\arctan(0.8/0.6)=\arctan(4/3)\approx0.9273\,\mathrm{rad}$. Taking $\phi=-\alpha$ gives

The starting displacement is not the amplitude when the initial velocity is nonzero. Using $\arccos(0.6)$ alone would give a positive phase and the wrong initial velocity. Phases differing by $2\pi$ describe the same motion.

4. Select the first future crossing

Equilibrium means $x=0$. For $x=A\cos(\omega t+\phi)$ with $A>0$, cosine is zero whenever $\omega t+\phi=\pi/2+n\pi$, where $n$ is any integer. Here $\omega=2.0\,\mathrm{rad/s}$ and $\phi=-\alpha$, so, with $t$ in seconds,

Choose the smallest $t>0$, not just any root. Here $n=-1$ gives a negative time; $n=0$ gives the first future crossing:

At this crossing,

The block is moving left. Its first right-hand turning point occurs earlier, at $t=\alpha/2\approx0.464\,\mathrm s$; that matches the initial direction we predicted.

5. Check what energy can tell you

For this ideal system, mechanical energy stays constant:

At a turning point, the block is momentarily at rest, so all the energy is stored in the spring. Thus $A=\sqrt{2E/k}=0.050\,\mathrm m$. At equilibrium, spring potential energy is zero and $E=\tfrac12mv^2$, giving $|v|=\sqrt{2E/m}=0.100\,\mathrm{m/s}$. Because energy contains $v^2$, it gives the speed but not the direction. The motion equation tells us that the first crossing is leftward and occurs at $t\approx1.249\,\mathrm s$.

6. Try two independent changes

Question 1 — Reverse the initial velocity.

Keep the original position, mass and spring stiffness, but set $v_0=-0.080\,\mathrm{m/s}$. Predict whether amplitude and first-crossing time change, then calculate the first crossing time and its signed velocity.

Hint: change the sine coefficient.

Only $D=v_0/\omega$ changes sign. Check both the sine and cosine of the new phase.

Solution: the same amplitude, a different starting direction.

The new motion is

$$\begin{aligned}x&=0.030\cos(2t)\\&\quad-0.040\sin(2t) \\&=0.050\cos(2t+\alpha).\end{aligned}$$

Amplitude remains $0.050\,\mathrm m$, period remains $\pi\,\mathrm s$, and energy remains $0.00250\,\mathrm J$. The block now moves directly toward equilibrium:

$$t_{\mathrm{first}}=\frac{\pi/2-\alpha}{2} \approx0.3218\,\mathrm s.$$

It crosses with $v=-0.100\,\mathrm{m/s}$. Equal energy and amplitude do not imply equal crossing times.

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Work through the starting point with a tutor

Start with an oscillation question you’re stuck on, or ask to review what position, velocity and phase mean. Your tutor can help you choose the equilibrium reference, interpret the initial direction, build the two initial-value equations and check the first event against a sketch. Then try a changed initial state with guidance where needed. Your current course outline can help choose the appropriate algebra or calculus approach. Sessions are currently offered in English. Topic support and availability are confirmed before booking.