Mistake Master
Mistake Master
Motion that repeats has a period: the time $T$ for one full cycle. Flip that and you get the frequency, $f = 1/T$, measured in hertz. For simple harmonic motion the period is set entirely by the system, not by how big the oscillation is: a spring's period uses $m$ and $k$, and a pendulum's uses $L$ and $g$. Amplitude shows up nowhere in either formula, which is the first place to go wrong.
A periodic motion repeats itself: position, velocity, and acceleration all return to the same values after a fixed time. That time is the period $T$. Flip it over and you get the frequency $f$, the number of cycles per second.
$T$ is measured in seconds, the time for one full cycle. $f$ is measured in hertz (Hz), or cycles per second: a frequency of 2 Hz means two complete cycles every second. The two are inverses of each other:
$$T = \dfrac{1}{f} \qquad f = \dfrac{1}{T}$$A common mix-up is to say something like "the frequency is half a second." Frequency is a rate (cycles per second), not a time, so the units alone rule it out. If you write a frequency in seconds, you've crossed wires with the period.
Once you know $T$, you can answer anything time-related about the motion: how many cycles occur in a given time, and when the block reaches each turning point and equilibrium crossing. The next question is what controls $T$. For SHM the answer comes in two clean formulas.
Topic 7.1's rule $F = -kx$ sets the stage. Two systems follow it: a block on an ideal spring, and a pendulum at small angles. Each one comes with its own period formula:
$$T_s = 2\pi\sqrt{\dfrac{m}{k}} \qquad T_p = 2\pi\sqrt{\dfrac{L}{g}}$$$T_s$ is for a spring-block oscillator: $m$ is the block's mass, $k$ is the spring's stiffness. $T_p$ is for a pendulum: $L$ is the string's length, $g$ is the strength of gravity. Both formulas have the same shape: $2\pi$ times the square root of an inertia variable on top and a restoring variable on bottom. Neither one has the amplitude in it.
A block of mass $m = 1$ kg sits on a frictionless surface, attached to a spring with $k = 4$ N/m. Pull it to $A = 1$ m and let go. Find the period and frequency.
Spring oscillator, so $T_s = 2\pi\sqrt{m/k}$.
$T_s = 2\pi\sqrt{1/4} = 2\pi \cdot \tfrac{1}{2} = \pi$ s.
$f = 1/T = 1/\pi \approx 0.32$ Hz.
A pendulum of length $L = 1$ m swings through a small angle near Earth's surface, where $g = 10$ N/kg. The bob has mass $0.5$ kg. Find the period and frequency.
Pendulum, so $T_p = 2\pi\sqrt{L/g}$.
$T_p = 2\pi\sqrt{1/10} = 2\pi \cdot \sqrt{0.1} \approx 1.99$ s.
$f = 1/T \approx 0.50$ Hz.
The square root softens the scaling. Doubling $m$ does not double the period; it multiplies it by $\sqrt{2} \approx 1.41$. To actually double $T$, you have to quadruple $m$. The same rule runs in reverse on the denominator: quadrupling $k$ divides $T$ by 2, not by 4.
Why doesn't amplitude appear? A larger swing means more distance per cycle, but the restoring force at the far endpoints is also bigger, so the block accelerates faster. The two effects scale together and cancel exactly. One round-trip takes the same time, regardless of how wide that round-trip is.
Three places where intuition pulls in the wrong direction. Each one is a version of the same root mistake: assuming a linear scaling where the formula has either no dependence at all, or a square root.
The guess feels right: more distance to travel should mean more time. The catch is that the speed scales up too. At a larger amplitude the restoring force at the turning point is larger, so the block leaves the turn moving faster. The maximum speed scales with $A$, and the distance to cover also scales with $A$. The two effects divide out and the period stays put.
Fix. Amplitude does not appear in $T_s = 2\pi\sqrt{m/k}$. It changes the energy ($U_s = \tfrac{1}{2}kA^2$) and the maximum speed, but not the cycle time. If you find yourself writing $A$ into a period formula, stop: it is not there.
Linear thinking left over from kinematics, where doubling the distance under constant acceleration does double the time. SHM is different: $T \propto \sqrt{m}$, not $m$. A factor of two in mass gives a factor of $\sqrt{2} \approx 1.41$ in the period, not a factor of two. To actually double $T$, you have to quadruple $m$.
Fix. Read the square root in the formula and respect it. When a variable under the square root changes by a factor $r$, the period changes by $\sqrt{r}$. The same rule runs for $k$ (and for $L$ and $g$ in the pendulum), so quadrupling $k$ halves $T$, not quarters it.
More mass should be harder to move, right? It is, but the formula tells a different story: the bob's mass is not in it. Here is why. Gravity pulls the bob with a force $mg$, so heavier bobs feel a larger restoring force at any given angle. But heavier bobs also have more inertia: pushing them around takes more force for the same speed-up. Both effects scale with $m$, and they cancel exactly.
Fix. A 5-kg bob and a 0.5-kg bob, hung from identical strings, tick at the same rate. Memorize the formula and trust it: $T_p = 2\pi\sqrt{L/g}$, with no $m$ in sight.
Ten scenarios, one trap each. Each scenario picks an oscillator and changes one variable; pick the right outcome from four choices. The applet tells you whether you got it right and, if not, what went wrong. Your progress saves as you go, so you can leave and come back.