Midterm 2: Example Solutions

 

(1a) In the limit where the oscillations are small, we can make the following approximations:

 

(i)           Tension in strings is the same as in equilibrium (2Mg and Mg for upper and lower strings respectively)

(ii)         Motions are nearly horizontal

(iii)       For any angle q to the vertical, sinq ~ tanq ~ q

 

è Force exerted by upper string on mass A = – 2MgY_A/L

Force exerted by lower string on mass A = – Mg(Y_A – Y_B )/L

Force exerted by lower string on mass B = + Mg(Y_A – Y_B )/L

 

Equations of motion: d²Y_A/dt² = (g/L)(Y_B – 3Y_A)

d²Y_B/dt² = (g/L)(Y_A – Y_B )

 

 

(b) Normal coordinates are of the form Y_N = Y_A + rY_B

 

for which d²Y_N/dt² = d²Y_A/dt² + r d²Y_B/dt²

= (g/L)(Y_B – 3Y_A + rY_A – rY_B)

= (g/L)([r – 3] Y_A + [1– r] Y_B)

 

The right-hand-side is a multiple of Y_N (in fact, [g/L][r – 3] times Y_N) provided (1 – r)/(r – 3) = r.

 

This requires 1 – r = r² – 3r <==> r² – 2r –1 = 0 <==> r = 1 ± Ö2

 

 

Normal coordinates are Y₁ = Y_A + (1 + Ö2) Y_B

Y₂ = Y_A + (1 – Ö2) Y_B

 

Equation of motion is then d²Y_N/dt² = (g/L)[–2 ±Ö2] Y_N

 

Frequencies are Ö[(g/L)(2 +Ö2)]

 

 

(c) Normal modes:

 

Y₁ = 0 ==> Y_A = – (1 + Ö2) Y_B

 

Antisymmetric mode in which masses move in opposite

directions.

 

Amplitude of mass A ~ 2.4 times amplitude of mass B

 

Frequency = Ö[(g/L)(2 + Ö2)]

 

 

 

Y₂ = 0 ==> Y_A = (Ö2 – 1) Y_B

 

Symmetric mode in which masses move in same directions

Amplitude of mass A ~ 0.4 times amplitude of mass B

 

Frequency = Ö[(g/L)(2 –Ö2)]

 

 

 

 

(2a) Equation of motion: d²Y/dt² + G dY/dt + w₀² Y = F₀ cos w₀t

where w₀ = Ö(K/M)

 

(b) The transient solutions to the equation die away on a timescale 1/G, leaving us with the steady-state solution for time, t > few x 1/G

 

(c) Try solution of form Y = Y₀ sin w₀t

(for which dY/dt = w₀Y₀ cos w₀t and d²Y/dt² = –w₀² Y₀ sin w₀t )

 

Substituting into equation of motion, we find that the first and third terms on the left-hand-side cancel, leaving us with

 

Gw₀Y₀ cos w₀t = F₀ cos w₀t, which is satisfied for all t if Y₀ = F₀/(Gw₀)

 

 

(2d) Velocity, dY/dt = w₀Y₀ cos w₀t

 

Power input = F dY/dt

= F₀ Y₀ w₀ cos²w₀t = (F₀²/G) cos²w₀t = GY₀² w₀² cos²w₀t

 

(Time average = ½ F₀ w₀Y₀ = ½ (F₀²/G) = ½ GY₀² w₀² )

 

The energy goes into heating the viscous liquid

 

 

(3) Key points:

 

Hard tiled walls have very high impedance, because they move very little in response to a force (è Z = force/velocity is small)

 

Hence, sound waves in air are reflected efficiently when they hit the walls, which act as high impedance dashpots:

 

R = (Z_air – Z_wall) / (Z_air + Z_wall) ~ –1 è reflected power ~ 100% and absorbed power is very small.

 

Similarly, water has a much higher impedance than air. Recall that for sound waves Z = Ö(gp₀r₀). For water, g = dlnp /dlnr is very large, water being nearly incompressible, and the density r₀ is also much larger than for air.

 

Hence the water is very reflective as well: reflected power ~ 100% and transmitted power is very small.

 

Bottom line: sound waves created in the air by excited screaming children bounce around for a long time with very little energy loss è it’s very nosy in an indoor pool.

 

Underwater, however, the sounds are much fainter because of the poor transmission of the air/water interface.