Difference between revisions of "Tugas rungekutta MFS"
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| − | [[File: | + | [[File:soalrunge.png]] |
'''Koding''' | '''Koding''' | ||
| − | + | import numpy as np | |
m = eval(input('massa benda: ')) | m = eval(input('massa benda: ')) | ||
| + | |||
g = eval(input('percepatan gravitasi: ')) | g = eval(input('percepatan gravitasi: ')) | ||
| + | |||
u = eval(input('koefisien gesek bidang: ')) | u = eval(input('koefisien gesek bidang: ')) | ||
| + | |||
cd = eval(input('koefisien gesek udara: ')) | cd = eval(input('koefisien gesek udara: ')) | ||
| + | |||
teta = eval(input('sudut kemiringan bukit: ')) * (np.pi / 180) | teta = eval(input('sudut kemiringan bukit: ')) * (np.pi / 180) | ||
| + | |||
sint = np.sin(teta) | sint = np.sin(teta) | ||
| + | |||
cost = np.cos(teta) | cost = np.cos(teta) | ||
| + | |||
v0 = 0 | v0 = 0 | ||
| + | |||
t0 = 0 | t0 = 0 | ||
| + | |||
dt = 1 | dt = 1 | ||
| + | |||
error = 100 | error = 100 | ||
| + | |||
tries = [] | tries = [] | ||
| + | |||
p = g * (sint - u * cost) | p = g * (sint - u * cost) | ||
| + | |||
q = cd / m | q = cd / m | ||
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print(f'kecepatan maksimum yang dapat dicapai adalah {round(v1, 2)} m/s dengan waktu {t1 + 1} s') | print(f'kecepatan maksimum yang dapat dicapai adalah {round(v1, 2)} m/s dengan waktu {t1 + 1} s') | ||
| − | + | ||
| + | [[File:hasilmfs.png]] | ||
| + | |||
| + | '''Video''' | ||
| + | |||
| + | [[File:Videorugekuttamfs.mkv]] | ||
Latest revision as of 16:02, 6 November 2019
Koding
import numpy as np
m = eval(input('massa benda: '))
g = eval(input('percepatan gravitasi: '))
u = eval(input('koefisien gesek bidang: '))
cd = eval(input('koefisien gesek udara: '))
teta = eval(input('sudut kemiringan bukit: ')) * (np.pi / 180)
sint = np.sin(teta)
cost = np.cos(teta)
v0 = 0
t0 = 0
dt = 1
error = 100
tries = []
p = g * (sint - u * cost)
q = cd / m
def dvdt(t0, v0):
return p - q * (v0 ** 1.5)
while error > 0.005:
k1 = dvdt(t0, v0) k2 = dvdt(t0 + 0.5 * dt, v0 + 0.5 * dt * k1) k3 = dvdt(t0 + 0.5 * dt, v0 + 0.5 * dt * k2) k4 = dvdt(t0 + dt, v0 + dt * k3) v1 = v0 + (1.0 / 6.0) * (k1 + 2 * k2 + 2 * k3 + k4) t0 = t0 + dt error = ((v1 - v0) / v1) * 100 v0 = v1 tries.append(v1)
t1 = len(tries)
print(f'kecepatan maksimum yang dapat dicapai adalah {round(v1, 2)} m/s dengan waktu {t1 + 1} s')
Video
