Fixed an incorrect data point

data/advanced_physics/seed_dataset.json

@@ -8163,7 +8163,7 @@
   },
   {
     "question": " An automobile with a mass of $1000 \\mathrm{~kg}$, including passengers, settles $1.0 \\mathrm{~cm}$ closer to the road for every additional $100 \\mathrm{~kg}$ of passengers. It is driven with a constant horizontal component of speed $20 \\mathrm{~km} / \\mathrm{h}$ over a washboard road with sinusoidal bumps. The amplitude and wavelength of the sine curve are $5.0 \\mathrm{~cm}$ and $20 \\mathrm{~cm}$, respectively. The distance between the front and back wheels is $2.4 \\mathrm{~m}$. Find the amplitude of oscillation of the automobile, assuming it moves vertically as an undamped driven harmonic oscillator. Neglect the mass of the wheels and springs and assume that the wheels are always in contact with the road.\r\n",
-    "final_answer": "-0.16 mm",
+    "final_answer": "0.16 mm",
     "rationale": "import sympy as sp\n\n# Define symbols for parameters\nm = sp.symbols('m', real=True, positive=True)       # mass of car\nk = sp.symbols('k', real=True, positive=True)       # effective suspension stiffness\nA = sp.symbols('A', real=True)                      # amplitude of road bump\nv = sp.symbols('v', real=True, positive=True)       # horizontal speed\nlam = sp.symbols('lam', real=True, positive=True)   # wavelength of bumps\nL = sp.symbols('L', real=True, positive=True)       # wheelbase (distance between front and rear wheels)\n\n# For a car with two springs acting on it, we assume the effective vertical motion (bounce mode) is governed by the average of the base displacements.\n# The front wheel sees a displacement: A*sin(2*pi*(x + L/2)/lam) and the rear wheel sees A*sin(2*pi*(x - L/2)/lam).\n# The effective forcing on the car body (its base excitation) is the average:\n#   y_b = 1/2*(A*sin(2*pi*(x+L/2)/lam) + A*sin(2*pi*(x-L/2)/lam))\n# Using the trigonometric identity, this equals A*cos(pi*L/lam)*sin(2*pi*x/lam).\n# Replacing x by v*t gives a sinusoidal forcing of angular frequency omega = 2*pi*(v/lam) and effective amplitude A_eff = A*cos(pi*L/lam).\n\n# The equation for undamped base excitation of a spring-mass system (with effective stiffness k and mass m) is:\n#         m*y'' + k*(y - y_b) = 0\n# The steady-state response for y is given by y = H(omega)*y_b, with transfer function\n#         H(omega) = -omega_n^2/(omega^2 - omega_n^2),   where omega_n^2 = k/m.\n# This negative sign indicates that at high forcing frequencies the body oscillates out of phase with the road.\n\nomega_n = sp.sqrt(k/m)                      # natural frequency\nomega = 2*sp.pi * (v/lam)                     # excitation (angular) frequency\n\n# Effective amplitude of the base excitation (from averaging the front and rear wheels):\nA_eff = A * sp.cos(sp.pi * L/lam)\n\n# The steady-state response amplitude for the car body (vertical bounce) is then (including the phase inversion):\nX = - (omega_n**2)/(omega**2 - omega_n**2) * A_eff\n\n# Assign numerical values based on the problem statement:\n# - Car mass (including passengers): 1000 kg\n# - The car settles 1.0 cm (0.01 m) closer to the road for every 100 kg extra, so an extra 100 kg (extra weight 980 N) produces 0.01 m deflection.\n#   Thus the effective suspension stiffness of the car as a whole is: k = 980 N/0.01 m = 98000 N/m.\n# - Road bump amplitude A: 5.0 cm = 0.05 m\n# - Road bump wavelength lam: 20 cm = 0.20 m\n# - Car speed v: 20 km/h = 20/3.6 m/s\n# - Wheelbase L: 2.4 m\n\nnum_m = 1000         # kg\nnum_k = 98000        # N/m\nnum_A = 0.05         # m\nnum_v = 20/3.6       # m/s\nnum_lam = 0.20       # m\nnum_L = 2.4          # m\n\n# Substitute numerical values into the expression for X\nX_sub = X.subs({m: num_m, k: num_k, A: num_A, v: num_v, lam: num_lam, L: num_L})\n\n# Evaluate the numerical value\nX_val = sp.N(X_sub)\n\n# The result X_val is the amplitude of vertical oscillation (in meters) of the car body. Since the expected answer is given in millimeters and includes the phase inversion,\n# multiply by 1000 to convert meters to millimeters.\nresult = X_val * 1000  # result in mm\n\n# Expected result is approximately -0.16 mm (negative sign indicates that the car oscillates out-of-phase with the bumps).\n\nunit=\"mm\"\nprint(f\"{result} {unit}\")",
     "metadata": {
       "name": "SciBench",
@@ -9846,4 +9846,4 @@
       ]
     }
   }
-]
+]