IB Physics C.1: Simple harmonic motion worked solutions

Amplitude is measured from the equilibrium level: half the range, (46-22)/2=12\,\mathrm{mm}. Highest to lowest is half a cycle, so T=0.80\,\mathrm{s} and \omega=2\pi/0.80=7.85\,\mathrm{rad\,s^{-1}}. A takes the whole range, C takes 0.40\,\mathrm{s} as the period, D gives the mid-height above the floor.

SHM needs both conditions: F proportional to x and directed towards the fixed point, so F=-kx. A points the right way but is not proportional. B is proportional but pushes the body away. D points the right way but its size does not depend on x (a marble rolling to and fro in a V-shaped groove behaves like this).

0.40\,\mathrm{s} is T/4. A quarter of a cycle after leaving an extreme, the bob passes through equilibrium: x=0, so a=-\omega^2x=0, and the speed is greatest. B describes t=0 and t=0.80\,\mathrm{s}.

Equal periods need equal l/g: l_E=0.40\times9.8/1.6=2.45\,\mathrm{m}. B uses the square root of the g ratio, D its square, A inverts it.

T=2\pi\sqrt{m/k}, so T\propto\sqrt m for the same spring: 0.60\times\sqrt{2.25}=0.60\times1.5=0.90\,\mathrm{s}. C makes T proportional to m; A divides instead of multiplying.

At equilibrium the speed is greatest, so E_k is greatest and the stored (elastic) energy least. At the extremes the tip is momentarily at rest (not A). Without damping the total is constant (not B, not D).

\omega=2\pi(260)=1634\,\mathrm{rad\,s^{-1}}; with x_0=8.0\times10^{-4}\,\mathrm{m}: v_{\max}=\omega x_0=1.31\,\mathrm{m\,s^{-1}} (A1), a_{\max}=\omega^2x_0=2.13\times10^{3}\,\mathrm{m\,s^{-2}} (A1), more than 200 times g. Leaving the amplitude in mm gives answers 1000 times too big.

At release v=0, so E_k=0. The speed peaks twice in each cycle (passing equilibrium in each direction), so E_k repeats every T/2=0.25\,\mathrm{s}. E_k=\tfrac12mv^2 cannot be negative (D); C has one peak per cycle; A starts at the wrong place.

In one cycle the particle travels 4x_0=0.12\,\mathrm{m} in T=0.20\,\mathrm{s}: average speed 0.60\,\mathrm{m\,s^{-1}}. A is the mean velocity (back where it started), B counts only 2x_0, D is the maximum speed 2\pi fx_0.

Gradient =-3.2/0.040=-80\,\mathrm{s^{-2}}, and a=-\omega^2x, so \omega^2=80 and \omega=8.94\,\mathrm{rad\,s^{-1}} (M1). f=\omega/2\pi=1.42\,\mathrm{Hz} (A1). Change the sign before taking the root, and do not stop at \omega.

Its displacement is now x=\tfrac34x_0, not \tfrac14x_0. E_p/E_T=(x/x_0)^2=\tfrac{9}{16}, so E_k/E_T=\tfrac{7}{16}. D uses x=\tfrac14x_0; C is the potential fraction; A squares the distance moved.

(a) T_1=12.4/20=0.620\,\mathrm{s} (M1). k=4\pi^2m/T^2=4\pi^2(0.200)/0.620^2=20.5\,\mathrm{N\,m^{-1}} (A1). (b) Same spring, so m\propto T^2: m_2=0.200(0.850/0.620)^2=0.376\,\mathrm{kg} (M1); stone =0.376-0.200=0.176\,\mathrm{kg} (A1). (c) Gravity only sets where the equilibrium position is; the period depends on m and k alone (A1). Timing 20 oscillations, not one, cuts the reaction-time uncertainty in T by a factor of 20.

(a) One surface is 2x higher than the other. The extra water in that column is not balanced by water opposite, so its weight drives the whole column back (M1). That weight is \rho A(2x)g, proportional to x, and it acts towards the level position (A1). (b) Mass moved =\rho AL. \rho AL\,a=-2\rho Agx (M1), so a=-(2g/L)x (A1). (c) \omega^2=2g/L: T=2\pi\sqrt{L/2g}=2\pi\sqrt{0.62/19.6}=1.12\,\mathrm{s} (A1). (d) \rho cancels, so the period is unchanged (A1). No pressure argument is needed: the unbalanced weight is enough.

(a) Moving the glider by x stretches one spring by an extra x and lets the other shorten by x. Each change of tension points back towards equilibrium (M1), so F=-30x-50x=-80x (A1). (b) T=2\pi\sqrt{0.80/80}=0.628\,\mathrm{s} (A1). (c) End to end, both springs carry the same force and their extensions add: 1/k=1/30+1/50, k=18.75\,\mathrm{N\,m^{-1}} (M1); T=2\pi\sqrt{0.80/18.75}=1.30\,\mathrm{s} (A1). Adding the periods, or using k=80 again, are the usual errors.

(a) \omega=2\pi(4.0)=25.1\,\mathrm{rad\,s^{-1}} (M1); x_0=v_{\max}/\omega=0.45/25.1=0.0179\,\mathrm{m} (A1). (b) Starting at the top: x=x_0\cos\omega t=0.018\cos(25t) (A1). (c) At the top the table accelerates downwards at \omega^2x_0=\omega v_{\max}=25.1(0.45)=11.3\,\mathrm{m\,s^{-2}} (M1). Gravity alone can give the coin only 9.8\,\mathrm{m\,s^{-2}} downwards, and the table cannot pull it down, so the coin loses contact (A1).

(a) Half the separation of the extremes: x_0=8.0\,\mathrm{cm} (A1). (b) 3.0\,\mathrm{cm} from a mark is x=5.0\,\mathrm{cm} from equilibrium (M1). \omega=2\pi(2.5)=15.7\,\mathrm{rad\,s^{-1}}; v=\omega\sqrt{x_0^2-x^2}=15.7\sqrt{0.080^2-0.050^2}=0.98\,\mathrm{m\,s^{-1}} (A1). (c) \omega^2x=15.7^2(0.050)=12.3\,\mathrm{m\,s^{-2}}, towards equilibrium (A1). (d) 1-(5.0/8.0)^2=39/64=0.61 (A1).

(a) v is the gradient of the x–t graph: zero at t=0, then negative as x falls, with peak size \omega x_0=\pi(0.050)=0.157\,\mathrm{m\,s^{-1}} at t=0.50\,\mathrm{s} (M1 shape and phase, A1 peak). a=-\omega^2x: the x graph inverted, peak size \pi^2(0.050)=0.49\,\mathrm{m\,s^{-2}} (A1). (b) a is always opposite to x: \pi rad (A1). (c) Speed is greatest at x=0, first at t=T/4=0.50\,\mathrm{s} (A1).

(a) E_p peaks at both extremes, twice per cycle, so T=2(0.25)=0.50\,\mathrm{s} (A1). (b) \omega=2\pi/0.50=12.6\,\mathrm{rad\,s^{-1}} (M1); x_0=\sqrt{\frac{2E_T}{m\omega^2}}=\sqrt{2(0.060)/(0.30\times12.6^2)}=0.0503\,\mathrm{m} (A1). (c) v_{\max}=\omega x_0=0.632\,\mathrm{m\,s^{-1}} (A1); check \tfrac12(0.30)(0.632)^2=0.060\,\mathrm{J}. (d) E_k=0.060-E_p: also between 0 and 0.060\,\mathrm{J}, period 0.25\,\mathrm{s}, maximum where E_p is zero (A1).

(a) E_k=E_T-E_p: the same parabola turned upside down, 0.090\,\mathrm{J} at the centre, zero at the extremes (A1). (b) E_T=\tfrac12kx_0^2: k=2(0.090)/0.060^2=50\,\mathrm{N\,m^{-1}} (A1). (c) T=2\pi\sqrt{m/k}=2\pi\sqrt{0.50/50} (M1) =0.628\,\mathrm{s} (A1). (d) E_p=\tfrac12E_T where (x/x_0)^2=\tfrac12: x=\pm0.060/\sqrt2=\pm0.042\,\mathrm{m}; each point is passed twice per cycle, four events in all (A1).

(a) For the spring, F=-kx; for SHM, F=ma=-m\omega^2x (M1). Equating for any x: m\omega^2=k, so E_T=\tfrac12kx_0^2 (A1). (b) E_T=\tfrac12(200)(0.040)^2=0.16\,\mathrm{J}=\tfrac12(0.50)v_{\max}^2, v_{\max}=0.80\,\mathrm{m\,s^{-1}} (A1). (c) E_T=\tfrac12kx_0^2 depends only on k and x_0: it is the work done stretching the spring, 0.16\,\mathrm{J} in both cases (M1). The heavier block shares that energy with a smaller speed, v_{\max}=\sqrt{2(0.16)/2.0}=0.40\,\mathrm{m\,s^{-1}}, so the claim is wrong (A1).

(a) Floating: \rho Ad\,g=mg, d=m/(\rho A)=0.040/(1000\times3.0\times10^{-4})=0.133\,\mathrm{m} (A1). (b) At equilibrium F_b=F_g. Pushing down by y displaces an extra volume Ay, so the buoyancy grows by \rho Ayg (M1); the resultant is this extra upward force, opposite to y: F=-\rho Agy (A1). (c) a=-(\rho Ag/m)y, the form a=-\omega^2y with \omega^2=\rho Ag/m=73.5\,\mathrm{s^{-2}} (M1); T=2\pi\sqrt{m/(\rho Ag)}=0.733\,\mathrm{s} (A1). The same as 2\pi\sqrt{d/g}. (d) T\propto1/\sqrt\rho: 0.733\sqrt{1000/1030}=0.722\,\mathrm{s} (A1). (e) The force law holds only while part of the tube stays in the water and its top stays above it. The top is 0.18-0.133=0.047\,\mathrm{m} above the surface, the nearer limit, so x_0\le0.047\,\mathrm{m} (A1). Water drag also damps the real motion (C.4).

(a) \sin\phi=-0.030/0.060=-0.5, so \phi=7\pi/6 or 11\pi/6 (M1). The velocity \omega x_0\cos\phi must be positive, so \cos\phi>0: \phi=11\pi/6 (A1). (b) \omega=2\pi/0.90=6.98\,\mathrm{rad\,s^{-1}}. Equilibrium next when \omega t+\phi=2\pi (M1): t=(\pi/6)/6.98=0.075\,\mathrm{s} (A1). (c) v=\omega\sqrt{x_0^2-x^2}=6.98\sqrt{0.060^2-0.030^2}=0.363\,\mathrm{m\,s^{-1}} (A1). (d) \omega t+\phi=2\pi+\pi/2: t=(2\pi/3)/6.98=0.300\,\mathrm{s} (A1). A quick check: 0.075\,\mathrm{s}+T/4=0.075+0.225=0.300\,\mathrm{s}.

(a) e_0=mg/k=0.20(9.8)/25=0.0784\,\mathrm{m} (A1); \tfrac12ke_0^2=0.0768\,\mathrm{J} (A1). (b) Greatest stretch at the lowest point: \tfrac12ke^2=0.206, e=\sqrt{2(0.206)/25}=0.128\,\mathrm{m} (M1). Amplitude =0.128-0.0784=0.050\,\mathrm{m} (A1): displacement is measured from equilibrium, not from the natural length. (c) T=2\pi\sqrt{0.20/25}=0.562\,\mathrm{s} (A1). (d) E_T=\tfrac12kx_0^2=\tfrac12(25)(0.050)^2=0.03125\,\mathrm{J}, so v_{\max}=\sqrt{2(0.03125)/0.20}=0.559\,\mathrm{m\,s^{-1}} (A1). (e) At the top the extension is 0.0784-0.050=0.028\,\mathrm{m}, still positive, so some elastic energy remains; the energy that oscillates is shared between elastic, kinetic and gravitational stores (A1).

(a) The four springs share the displacement, so their forces add: k_{\text{eff}}=4(2.0\times10^4)=8.0\times10^4\,\mathrm{N\,m^{-1}} (M1); T=2\pi\sqrt{1200/(8.0\times10^4)}=0.770\,\mathrm{s} (A1). (b) Extra compression =300(9.8)/(8.0\times10^4)=0.0368\,\mathrm{m} (A1). T=2\pi\sqrt{1500/(8.0\times10^4)} (M1) =0.860\,\mathrm{s} (A1). (c) One bump per period: v=6.0/0.860=6.97\,\mathrm{m\,s^{-1}} (A1). Driving the car at its natural frequency builds up large oscillations, resonance (C.4); real shock absorbers damp it (A1).

(a) Radian mode: x=0.050\sin0.60=0.0282\,\mathrm{m} (A1); v=\omega x_0\cos(\omega t+\phi)=0.20\cos0.60=+0.165\,\mathrm{m\,s^{-1}} (A1). (b) 4.0\sqrt{0.050^2-0.0282^2}=0.165\,\mathrm{m\,s^{-1}}: the same speed (A1). (c) x=0 again when 4.0t+0.60=\pi (M1), t=(\pi-0.60)/4.0=0.635\,\mathrm{s} (A1). (d) E_T=\tfrac12m\omega^2x_0^2=\tfrac12(0.80)(16)(0.050)^2=0.016\,\mathrm{J} (A1); E_k=\tfrac12(0.80)(0.165)^2=0.0109\,\mathrm{J} (A1), the rest, 0.0051\,\mathrm{J}, is potential.

(a) T=2\pi\sqrt{0.75/9.8}=1.74\,\mathrm{s} (A1). (b) In the lift's frame the bob behaves as if gravity were g+a: the string must both support the bob and accelerate it, so the tension and the restoring component are larger (M1). g_{\text{eff}}=9.8+1.5=11.3\,\mathrm{m\,s^{-2}} (M1); T=2\pi\sqrt{0.75/11.3}=1.62\,\mathrm{s} (A1). (c) In free fall the bob and lift accelerate together at g; nothing pulls the bob back towards the lowest point relative to the lift. A moving bob is kept on the string's circle by the tension and goes round the pivot at constant speed; a bob at rest at an extreme stays there (A1).

(a) D.1: g=GM/R^2=6.67\times10^{-11}(6.42\times10^{23})/(3.39\times10^6)^2 (M1) =3.73\,\mathrm{N\,kg^{-1}} (A1). (b) T=2\pi\sqrt{0.500/3.73}=2.30\,\mathrm{s} (A1), against 1.42\,\mathrm{s} on Earth. (c) 0.800\,\mathrm{s} (A1): the spring's period depends on m and k only; weaker gravity just shortens the equilibrium stretch (A1). (d) g_{\text{eff}}=3.73+2.0=5.73\,\mathrm{m\,s^{-2}} (M1); T=2\pi\sqrt{0.500/5.73}=1.86\,\mathrm{s} (A1).

(a) C.2: f=v/\lambda=0.60/0.48=1.25\,\mathrm{Hz} (A1); T=0.80\,\mathrm{s} (A1). (b) One crest passes the buoy per period, so it is pushed up and down at the wave frequency (A1). The wave speed is the speed of the pattern; the water and the buoy only oscillate about a fixed level (A1). (c) v_{\max}=\omega x_0=2\pi(1.25)(0.025)=0.196\,\mathrm{m\,s^{-1}} (A1). (d) a_{\max}=\omega^2x_0=1.54\,\mathrm{m\,s^{-2}} (M1). This is less than g, so gravity can supply the downward acceleration at the top: the shell stays on (A1). (e) 0.36/0.48=0.75 of a wavelength, so \Delta\phi=0.75(2\pi)=3\pi/2 rad: the far buoy lags by 3\pi/2, which is the same as leading by \pi/2 (A1).

(a) In contact, the resultant force is the spring force -kx, proportional to the compression and directed back (M1). The trolley enters at equilibrium (spring unstressed) and leaves there after reversing: half a cycle (A1). (b) t=T/2=\pi\sqrt{m/k} (M1) =\pi\sqrt{0.60/150}=0.199\,\mathrm{s} (A1). (c) A.3: \tfrac12mv^2=\tfrac12kx^2 (M1), x=0.80\sqrt{0.60/150}=0.0506\,\mathrm{m} (A1). (d) A.2: velocity reverses, impulse magnitude =|\Delta p|=2mv=0.96\,\mathrm{N\,s} (A1); F_{\text{avg}}=0.96/0.199=4.83\,\mathrm{N} (A1). F_{\max}=kx=7.59\,\mathrm{N}; the average is 2/\pi of the peak, because the force rises and falls as a half sine (A1).

(a) A.2: \omega=45(2\pi)/60=4.71\,\mathrm{rad\,s^{-1}} (A1). (b) The peg accelerates towards the centre at \omega^2r (M1). The shadow shows only the component along the wall line; when the peg is at angle \theta from that line its shadow is at x=r\cos\theta and its acceleration component is -\omega^2r\cos\theta (M1). So a=-\omega^2x: proportional to x and towards the centre of the line, SHM with the turntable's \omega (A1). (c) x_0=0.12\,\mathrm{m} (A1); T=2\pi/\omega=1.33\,\mathrm{s} (A1); v_{\max}=\omega r=0.565\,\mathrm{m\,s^{-1}}, the peg's own speed, reached as the shadow crosses the centre (A1). (d) l=gT^2/4\pi^2=9.8(1.333)^2/4\pi^2=0.441\,\mathrm{m} (A1).