Corrected exercises: oscillations and simple harmonic motion (PHYS 101)

The amplitude is the largest elongation reached, measured from the equilibrium line and not from the bottom of the curve: the curve runs from $+6.0$ cm to $-6.0$ cm, so $A = 6.0$ cm and NOT 12.0 cm. The period is the time for one complete pattern, so from one maximum to the NEXT maximum: the first maximum is at $t = 0$, the second at $t = 0.80$ s, hence $T = 0.80$ s. Counting from a maximum to the following minimum would give $0.40$ s, which is half a period, and it is the single most common reading error on this graph.

The frequency is the number of complete cycles per second, $f = 1/T = 1/0.80 = 1.25$ Hz. The angular frequency counts the same motion in radians instead of cycles, and one cycle is $2\pi$ radians: $\omega = 2\pi f = 2\pi/T = 2\pi/0.80 \approx 7.85$ rad/s. Check the ratio: $\omega/f = 6.28$, which is $2\pi$. A value of $\omega$ that comes out close to $f$ means you wrote $\omega = 1/T$, and every later number will be wrong by a factor of $6.28$.

At $t = 0$ the sensor reads $x = +A$: the glider was held at maximum elongation and released from rest. The function that equals its own maximum at $t = 0$ is the cosine, so $x(t) = 6.0\cos(7.85\,t)$ with $x$ in centimetres. Test it: at $t = 0.20$ s, $7.85 \times 0.20 = 1.57$ rad, and $\cos(1.57) = 0$, which matches the graph crossing zero at $t = 0.20$ s. Had the glider been launched from the equilibrium point with a push, the graph would start at zero and the sine would be the right choice. The argument $\omega t$ is in RADIANS, so the calculator must be in radian mode.

Releasing it from further away changes the LAUNCH, not the SYSTEM. The spring constant and the mass are untouched, so $\omega$, $T$ and $f$ keep exactly the values found above: the new curve crosses zero at the same instants and has the same spacing between maxima. What changes is the amplitude, $A = 12.0$ cm, and everything built on it: the maximum speed $v_{max} = \omega A$ goes from $7.85 \times 0.060 = 0.47$ m/s to $7.85 \times 0.120 = 0.94$ m/s, the maximum acceleration doubles as well, and the total energy $\frac{1}{2}kA^{2}$ is multiplied by $2^{2} = 4$. This is the thread of the whole chapter: doubling $A$ doubles the speeds and quadruples the energy, and leaves the clock alone.

The spring constant is the SLOPE of the line, not one of the plotted points. Take two points far apart to keep the reading error small: $(0.05\ \text{m}; 1.2\ \text{N})$ and $(0.20\ \text{m}; 4.8\ \text{N})$. Then $k = \frac{4.8 - 1.2}{0.20 - 0.05} = \frac{3.6}{0.15} = 24$ N/m. The unit follows from the division, newtons per metre. Dividing a single point, $1.2/0.05$, gives the same 24 here only because the line passes through the origin, which is exactly what makes the law linear; on a graph with an offset that shortcut is wrong.

Two features. Proportional: the number $k$ does not change along the line, so pulling twice as far calls up twice the force. Opposite: the minus sign says the force points back towards $x = 0$ whichever side you are on. Together they are the ONLY condition needed for simple harmonic motion, and every oscillator in this chapter is checked against it. The graph shows the first feature directly, since the points lie on a straight line through the origin; the second is the experimental fact that the spring pulls when stretched and pushes when compressed.

The period of a mass on a spring is $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.60/24} = 2\pi\sqrt{0.025} = 2\pi \times 0.1581 \approx 0.99$ s, and $f = 1/T \approx 1.01$ Hz. Check the unit inside the root: kg divided by N/m is $\text{kg}\cdot\text{m}/\text{N}$, and one newton is $\text{kg}\cdot\text{m}/\text{s}^{2}$, so the fraction is s$^{2}$ and its square root is a time. If your answer comes out in the hundreds or the thousandths, you have inverted $m/k$.

At $x = 0.050$ m the restoring force is $F = kx = 24 \times 0.050 = 1.2$ N and the acceleration is $a = F/m = 1.2/0.60 = 2.0$ m/s$^{2}$. At $x = 0.100$ m, $F = 2.4$ N and $a = 4.0$ m/s$^{2}$: both double, because both are proportional to $x$. The period, on the other hand, is $2\pi\sqrt{m/k}$, an expression in which the amplitude does not appear at all: it stays $0.99$ s in both trials. Far from equilibrium the cart is pulled harder, so it travels its longer path faster, and the two effects cancel exactly. That exact cancellation is the property that makes a spring a clock.

$\omega = 2\pi/T = 2\pi/0.50 = 4\pi \approx 12.57$ rad/s.

The speed factor in $v = -\omega A\sin(\omega t)$ is largest when the sine equals $\pm 1$, so $v_{max} = \omega A = 12.57 \times 0.12 \approx 1.51$ m/s. The sine equals $\pm 1$ exactly when the cosine is zero, that is when $x = 0$: the cone is fastest as it passes through the equilibrium point. This is the position where the spring force is zero, which surprises students who expect the force and the speed to peak together; in fact the cone has been accelerated over the whole quarter cycle before, and equilibrium is where it has finished gathering speed.

Likewise $a_{max} = \omega^{2}A = (12.57)^{2} \times 0.12 = 157.9 \times 0.12 \approx 18.9$ m/s$^{2}$, reached when the cosine equals $\pm 1$, that is at $x = \pm A$, the two turning points. The speed there is ZERO. Velocity and acceleration reach their maxima a quarter of a cycle apart, never together, and at each turning point the cone is momentarily stopped while being pushed back the hardest it is pushed anywhere. Saying that the acceleration is zero because the cone has stopped is the classic error: if it were zero the cone would simply stay there.

For the speed the course gives $v = \omega\sqrt{A^{2}-x^{2}}$, which is the same three equations with the time eliminated: $v = 12.57\sqrt{0.12^{2}-0.06^{2}} = 12.57\sqrt{0.0108} = 12.57 \times 0.1039 \approx 1.31$ m/s, that is $0.87\,v_{max}$. For the acceleration, $a = \omega^{2}x = 157.9 \times 0.060 \approx 9.5$ m/s$^{2}$, exactly $a_{max}/2$. The two behave completely differently at the same point: the acceleration is proportional to $x$, so half the elongation gives half the acceleration, while the speed follows a square root and still holds $87\%$ of its maximum at half amplitude. An oscillator spends much more of its time near the ends of its path than a uniform-speed picture suggests, and the graph opposite shows why: the velocity curve is flat and broad around its peak, the position curve is not.

The peg moves on a circle at constant speed, so the formula sheet gives its speed along the tangent and its acceleration towards the centre:

$v = \omega R$ and $a = \frac{v^{2}}{R} = \frac{(\omega R)^{2}}{R} = \omega^{2}R$

The shadow keeps only the part of each vector that runs ALONG the screen; the part pointing at the screen is lost in the projection.

Speed: the peg's velocity lies along the screen when the peg is nearest to it or furthest from it. Its shadow is then at $x = 0$, so the shadow's largest speed is $v_{max} = \omega R$, at the centre of its path.

Acceleration: the centripetal acceleration lies along the screen when the peg is at the far left or the far right of the turntable. Its shadow is then at $x = \pm R$, so $a_{max} = \omega^{2}R$, at the two ends.

At those two ends the velocity points straight at the screen and the shadow stops; at the centre the acceleration points straight at the screen and the shadow has none. Parts b and c are recovered without any equation of motion.

At any angle, the part of the centripetal acceleration along the screen is $-\omega^{2}R\cos(\omega t) = -\omega^{2}x$: the shadow obeys the condition of simple harmonic motion exactly.

Dimensions: $[\omega R] = \text{s}^{-1}\cdot\text{m} = \text{m/s}$ and $[\omega^{2}R] = \text{s}^{-2}\cdot\text{m} = \text{m/s}^{2}$, a speed and an acceleration, since the radian carries no unit.

Limiting case: as $\omega \to 0$ the turntable stops, the shadow stands still, and both expressions go to zero, as they must. Direction of change: turning twice as fast doubles $v_{max}$ but multiplies $a_{max}$ by four.

Numbers: one turn in $2.0$ s gives $\omega = \frac{2\pi}{2.0} = \pi \approx 3.14$ rad/s.

$v_{max} = 3.14 \times 0.15 \approx 0.471$ m/s and $a_{max} = \pi^{2} \times 0.15 \approx 1.48$ m/s$^{2}$

The circle also says what the phase is: the angle of the peg when the stopwatch starts. Exercise 4 f uses it.

At the instant of release the block is at rest at $x = A$, so all the energy is elastic: $E = \frac{1}{2}kA^{2} = 0.5 \times 200 \times (0.100)^{2} = 1.00$ J. Since there is no friction, that number is fixed for the whole motion: the oscillator only ever moves this one joule back and forth between the two forms.

At $x = 0$ the spring is relaxed, so the whole joule is kinetic: $\frac{1}{2}mv_{max}^{2} = 1.00$ J gives $v_{max} = \sqrt{2 \times 1.00/0.50} = \sqrt{4} = 2.00$ m/s. Independent check through the clock: $\omega = \sqrt{k/m} = \sqrt{200/0.50} = \sqrt{400} = 20$ rad/s, and $v_{max} = \omega A = 20 \times 0.100 = 2.00$ m/s. The two routes agree, which is worth doing once in an exam because it catches a wrong $k$ or a mass entered in grams.

At $x = 0.050$ m the elastic part is $E_{p} = 0.5 \times 200 \times (0.050)^{2} = 0.25$ J, so the kinetic part is what is left, $1.00 - 0.25 = 0.75$ J. Then $v = \sqrt{2 \times 0.75/0.50} = \sqrt{3} \approx 1.73$ m/s. That is $87\%$ of the maximum speed, not $50\%$. The reason is in the squares: halving $x$ divides the POTENTIAL energy by four, not by two, so three quarters of the energy is still kinetic at half amplitude. Writing $v = v_{max}/2$ here costs the whole question and it is the most frequent mistake of the chapter.

Equal shares means each form holds $0.500$ J. From $\frac{1}{2}kx^{2} = 0.500$ we get $x^{2} = 2 \times 0.500/200 = 0.005$, so $x = 0.0707$ m, that is $7.07$ cm, which is $A/\sqrt{2} \approx 0.71\,A$. The speed there is $v = \sqrt{2 \times 0.500/0.50} = \sqrt{2} \approx 1.41$ m/s, exactly $v_{max}/\sqrt{2}$. Note that the halfway point in ENERGY sits at $71\%$ of the amplitude, not at $50\%$: the two parabolas of the figure cross well to the right of the middle, and reading that crossing correctly is the whole point of drawing them.

Nothing is lost to friction, so the energy at the release equals the energy at the turning point, where the block is momentarily at rest at $x = A$:

Multiply by $\frac{2}{k}$ and take the root: $A = \sqrt{x_{0}^{2} + \frac{m}{k}v_{0}^{2}}$.

Dimensions: $\frac{m}{k}$ is in $\text{s}^{2}$, as exercise 2 c showed, so $\frac{m}{k}v_{0}^{2}$ is in $\text{s}^{2}\cdot\text{m}^{2}/\text{s}^{2} = \text{m}^{2}$, like $x_{0}^{2}$: the two terms can be added and the root is a length.

Limiting case $v_{0} \to 0$: $A = x_{0}$, the release from rest of part a, where the release point IS the amplitude.

Limiting case $x_{0} \to 0$: $A = v_{0}\sqrt{\frac{m}{k}} = \frac{v_{0}}{\omega}$, that is $v_{0} = \omega A$: a launch from the centre gives the block its maximum speed, as part b says.

Direction of change: $A$ grows with the push and always exceeds $x_{0}$. The sign of $v_{0}$ does not enter, since it is squared: a push outwards or inwards gives the same amplitude.

$A = \sqrt{0.060^{2} + \frac{0.50}{200} \times 1.60^{2}} = \sqrt{0.0036 + 0.0064} = 0.100$ m

So $A = 10.0$ cm. The energy is $0.36 + 0.64 = 1.00$ J, the joule of part a: this is the same oscillation as the first trial, started at another point of its cycle.

Pitfall: taking $A = 6.0$ cm because the block was let go there. With a push, the release point is not a turning point.

At $t = 0$ the position equation gives $x_{0} = A\sin\varphi$:

The calculator's arcsine returns $\varphi = 0.644$ rad. But a sine takes the value $0.600$ twice per cycle, at $0.644$ rad and at $\pi - 0.644 = 2.498$ rad.

The velocity decides. At $t = 0$, $v_{0} = \omega A\cos\varphi$ must be NEGATIVE, so $\cos\varphi < 0$.

$\cos(0.644) = +0.800$ would send the block outwards at $+1.60$ m/s: rejected. $\cos(2.498) = -0.800$ gives $v_{0} = 20 \times 0.100 \times (-0.800) = -1.60$ m/s: kept.

So $\varphi = 2.50$ rad, about $143^{\circ}$, and $x = 0.100\sin(20t + 2.50)$ in metres.

The sine next vanishes when its argument reaches $\pi$:

$20t + 2.498 = \pi$, so $t = \frac{\pi - 2.498}{20} = \frac{0.644}{20} \approx 0.032$ s

Direction of change: at a constant $1.60$ m/s the $6.0$ cm would take $\frac{0.060}{1.60} = 0.0375$ s. The spring speeds the block up on its way in, so the true time must be shorter, and $0.032$ s is.

Pitfall: keeping the calculator's $0.644$ rad. That curve, dashed in the graph below, leaves AWAY from the centre and first reaches $x = 0$ at $0.125$ s, four times too late.

$T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.25/40} = 2\pi\sqrt{0.00625} = 2\pi \times 0.0791 \approx 0.497$ s.

Hanging at rest, the mass is in equilibrium: the spring force pulling up balances the weight pulling down, $kd = mg$, so $d = mg/k = 0.25 \times 9.8/40 = 0.0613$ m, about $6.1$ cm. Note what this measurement is worth in the laboratory: a ruler and a known mass give $k$ without any timing at all.

The period is $0.497$ s again, the very same number. Measure the elongation $x$ from the NEW resting position, the one $6.1$ cm below the unloaded end. At an elongation $x$ below it, the spring is stretched by $d + x$, so it pulls up with $k(d+x)$ while gravity pulls down with $mg = kd$. The resultant is $k(d+x) - kd = kx$, directed back towards the new centre. That is exactly the condition $F = -kx$ with the SAME $k$, so the same formula gives the same period. Gravity has done one thing only: it has moved the centre of the oscillation down by $d$. Note the shortcut it hands you, $T = 2\pi\sqrt{d/g} = 2\pi\sqrt{0.0613/9.8} \approx 0.497$ s, a period obtained from a ruler alone, with neither $k$ nor $m$.

At $4.0$ cm below the hanging position the spring is stretched by $0.0613 + 0.040 = 0.1013$ m, so it pulls up with $40 \times 0.1013 = 4.05$ N. The weight is $0.25 \times 9.8 = 2.45$ N downward. The resultant is $4.05 - 2.45 = 1.60$ N upward, and $k \times 0.040 = 1.60$ N: the two agree to the last digit, as part c predicted. The acceleration is $1.60/0.25 = 6.4$ m/s$^{2}$, which is also $\omega^{2}A = (k/m)A = 160 \times 0.040$. The practical conclusion for an exam: in a vertical spring problem, never carry the weight through the calculation. Measure $x$ from the hanging position and the weight has already been accounted for.

Side by side, both springs are stretched by the same amount $x$ and both pull on the block, so the forces add: $F = k_{1}x + k_{2}x = (k_{1}+k_{2})x$, and the effective constant is $k = 300 + 600 = 900$ N/m. Then $T = 2\pi\sqrt{2.0/900} = 2\pi \times 0.0471 \approx 0.296$ s.

End to end, the two springs carry the SAME force (each one pulls on the next), and their stretches add. A given force $F$ gives $F/300 + F/600 = F/200$ of total stretch, so the effective constant is $200$ N/m. Then $T = 2\pi\sqrt{2.0/200} = 2\pi \times 0.100 \approx 0.628$ s. Notice that the combined constant, $200$ N/m, is SMALLER than either spring on its own, which is the sign that tells you the reciprocal rule has been applied the right way round.

The ratio is $0.628/0.296 = 2.12$, and indeed $T \propto 1/\sqrt{k}$ gives $\sqrt{900/200} = \sqrt{4.5} = 2.12$. In words: end to end, the block can move a long way for a small force, because each spring gives a little and the two gifts add. The assembly is softer than either component, so the restoring force at a given elongation is weaker, so the return trip is slower. Side by side the two springs resist together at every elongation, the assembly is stiffer than either one, and the block is rushed back to the centre. The mass has not changed in either case, only the stiffness, and the period tracks the stiffness alone.

Think of the original spring as two halves mounted end to end, since that is exactly what it is. If each half has constant $k_{h}$, the reciprocal rule gives $\frac{1}{300} = \frac{1}{k_{h}} + \frac{1}{k_{h}} = \frac{2}{k_{h}}$, so $k_{h} = 600$ N/m: cutting a spring in half DOUBLES its stiffness, because the same pull now has to be taken up by half as many coils. The whole spring would give $T = 2\pi\sqrt{2.0/300} \approx 0.513$ s and the half spring gives $T = 2\pi\sqrt{2.0/600} \approx 0.363$ s, shorter by a factor $\sqrt{2}$. Doubling $k$ does not halve the period, it divides it by $1.41$, because $k$ sits under a square root.

$T = 2\pi\sqrt{0.60/9.8} = 2\pi\sqrt{0.0612} = 2\pi \times 0.2474 \approx 1.55$ s, and $f = 1/T \approx 0.64$ Hz. Note that the full period is the time for a there-and-back trip; the time from one extreme to the other is only $0.78$ s, and calling that the period is a standard way to lose the question.

Neither change touches the period. The mass does not appear in $T = 2\pi\sqrt{L/g}$ at all: a heavier bob is pulled back by a proportionally larger force, and needs a proportionally larger force to be accelerated the same way, so the two effects cancel exactly, as they do for objects falling freely. The angle does not appear either, as long as it stays small: going from $5^{\circ}$ to $10^{\circ}$ lengthens the true period by about $0.14\%$, which is two milliseconds here and which no stopwatch in the laboratory will show. Only $L$ and $g$ are left, which is why a pendulum is a clock and also why it can be used to measure $g$.

Solve the formula for $L$: from $T = 2\pi\sqrt{L/g}$, squaring gives $T^{2} = 4\pi^{2}L/g$, so $L = gT^{2}/(4\pi^{2}) = 9.8 \times 4.00/39.48 = 0.993$ m. The historical seconds pendulum, which ticks once per second and so has a period of two seconds, is almost exactly one metre long, and that near-coincidence was once proposed as the definition of the metre. Check the direction of the dependence before writing the answer: four times the length gives twice the period, so a length near one metre for a period near twice that of the $0.60$ m pendulum is consistent.

At $10^{\circ}$: $\theta = 0.1745$ rad and $\sin\theta = 0.1736$, so they differ by $0.5\%$. At $30^{\circ}$: $\theta = 0.5236$ rad and $\sin\theta = 0.5000$, a disagreement of $4.5\%$. The restoring force along the arc is $mg\sin\theta$ while the formula assumes $mg\theta$, so the formula slightly overestimates the pull at large angles and therefore UNDERESTIMATES the period: at $30^{\circ}$ the true period is about $1.7\%$ longer than $2\pi\sqrt{L/g}$, which a careful timing of twenty swings will detect. Keep releases under about $15^{\circ}$ in the laboratory. And note where the radians came from: the approximation is a statement about radians only, and a calculator left in degree mode gives $\sin(10) = 0.174$ against $\theta = 10$, a comparison that means nothing.

FALSE. The period of a mass-spring oscillator is $T = 2\pi\sqrt{m/k}$, an expression that contains neither the amplitude nor the way the oscillator was launched. Doubling $A$ doubles $v_{max} = \omega A$ and $a_{max} = \omega^{2}A$, and multiplies the energy $\frac{1}{2}kA^{2}$ by four, while the clock keeps time. Correction: the period is unchanged.

FALSE, and the second half is where it goes wrong. At $x = \pm A$ the velocity is indeed zero, for one instant. But $a = -\omega^{2}x$, so the acceleration is at its LARGEST magnitude exactly there. It has to be: if both were zero the object would stay at the end of its path and there would be no oscillation at all. Correction: at the turning points, $v = 0$ and $|a| = \omega^{2}A$. The confusion comes from everyday language, where stopping is associated with nothing happening.

TRUE. $T = 2\pi\sqrt{m/k}$, so $T$ is proportional to $\sqrt{m}$ and multiplying the mass by four multiplies the period by $\sqrt{4} = 2$. A numerical check: with $k = 100$ N/m, $m = 1.0$ kg gives $T = 0.628$ s and $m = 4.0$ kg gives $T = 1.257$ s. Beware of carrying this over to the pendulum, where the mass does not appear at all.