ralat
TRANSCRIPT
RALAT
A. Menghitung Panjang Gelombang Secara Eksperimen
∆ λn=| ∂ λn∂ x0 xn||∆ s|=|2
n||∆s|
Dimana :
∆ s=12NST Mistar=5×10−4m
1. Pada Tali Putih
Untuk m1 = 0,07 kg
Δ λ1=| ∂ λ1
∂ X0 Xn|Δ s=|2
n|Δ s=21
(0,0005 )=0,001m
KTPR ¿Δ λ1
λ1×100 %=0,001
0,20×100 %=0,5 %
Ketelitian¿100 %−KTPR=100 %−0,5 %=99,5 %
AB = 1 – log (Δ λ1
λ1) = 1 – log (
0.0010,20 ) = 1−¿(-2,30) = 3,30 ≈3 AB
Pelaporan:λ1± Δ λ1=(0,5±0,001 )m
λ1+Δ λ1= (50+1 )×10−3m=51×10−3m
λ1−Δ λ1=(50−1 )×10−3m=49×10−3m
Δ λ2=| ∂ λ2
∂ X0 X |=|2n|Δ s=2
2(0,0005 )=0,0005m
KTPR =Δ λ2
λ2×100 %=
0,00050,11
×100%=0,45%
Ketelitian = 100%−KTPR= 100% −¿ 0,45% = 99,55%
AB = 1 – log (Δ λ2
λ2) = 1 – log (
0.00050,11 ) = 1−¿(-2,34) = 3,34≈3 AB
Pelaporan:λ2± Δ λ2=(0,11±0,0005 )m
λ2+Δ λ2= (110+0,5 )×10−3m=110,5×10−3m
λ2−Δ λ2=(110−0,5 )×10−3m=109,5×10−3m
Δ λ3=| ∂ λ3
∂ X0 X |=|2n|Δ s=2
3(0,0005)=0,0003m
KTPR =Δ λ3
λ3×100 %=
0,00030,086
×100 %=0,35%
Ketelitian = 100%−KTPR= 100% −¿ 0,35% = 99, 65%
AB = 1 – log (Δ λ3
λ3)= 1 – log (
0.00030,086 )= 1−¿(-2,45) = 3,45≈3 AB
Pelaporan:λ3± Δ λ3=(0,086±0,0003 )m
λ3+Δ λ3=(86+0,3 )×10−3m=86,3×10−3m
λ3−Δ λ3=(86−0,3 )×10−3m=85,7×10−3m
Untuk m2 = 0,09 kg
Δ λ1=| ∂ λ1
∂ X0 X |=|2n|Δs=2
1( 0,0005 )=0,001m
KTPR ¿Δ λ1
λ1×100 %=0,001
0,54×100 %=0,18 %
Ketelitian¿100 %−KTPR=100 %−0,18 %=99,82 %
AB = 1 – log (Δ λ1
λ1) = 1 – log (
0.0010,54 ) = 1−¿(-2,73) = 3,73≈4 AB
Pelaporan:λ1± Δ λ1=(0,54±0,001 )m
λ1+Δ λ1= (0,54+0,001 )m=0,54m
λ1−Δ λ1=(0,54−0,001 )m=0,539m
Δ λ2=| ∂ λ2
∂ X0 X |=|2n|Δs=2
2(0,0005 )=0,0005m
KTPR =Δ λ2
λ2×100 %=
0,00050,28
×100 %=0,17 %
Ketelitian = 100%−KTPR= 100% −¿ 0,17% = 99,83%
AB = 1 – log (Δ λ2
λ2) = 1 – log (
0.00050,28 ) = 1−¿(-2,74) = 3,74≈4 AB
Pelaporan:λ2± Δ λ2=(0,28±0,0005 )m
λ2+Δ λ2= (0,28+0,0005 )m=0,2805m
λ2−Δ λ2=(0,28−0,0005 )m=0,2795m
Δ λ3=| ∂ λ3
∂ X0 X |=|2n|Δs=2
3(0,0005 )=0,0003m
KTPR =Δ λ3
λ3×100 %=
0,00030,14
×100 %=0,21%
Ketelitian = 100%−KTPR= 100% −¿ 0,21% = 99, 79%
AB = 1 – log (Δ λ3
λ3)= 1 – log (
0.00030,14 )= 1−¿(-2,67) = 3,67≈4 AB
Pelaporan:λ3± Δ λ3=(0,14±0,0003 )m
λ3+Δ λ3=(0,14+0,0003 )m=0,1403m
λ3+Δ λ3=(0,14+0,0003 )m=0,1397m
Untuk m3 = 0,11 kg
Δ λ1=| ∂ λ1
∂ X0 X |=|2n|Δs=2
1( 0,0005 )=0,001m
KTPR ¿Δ λ1
λ1×100 %=0,001
0,22×100 %=0,45 %
Ketelitian¿100 %−KTPR=100 %−0,45 %=99,55 %
AB = 1 – log (Δ λ1
λ1) = 1 – log (
0.0010,22 ) = 1−¿(-2,34) = 3,34≈3 AB
Pelaporan:λ1± Δ λ1=(0,22±0,001 )m
λ1+Δ λ1= (0,22+0,001 )m=0,221m
λ1−Δ λ1=(0,22−0,001 )m=0,219m
Δ λ2=| ∂ λ2
∂ X0 X |=|2n|Δs=2
2(0,0005 )=0,0005m
KTPR =Δ λ2
λ2×100 %=
0,00050,15
×100%=0,33%
Ketelitian = 100%−KTPR= 100% −¿ 0,33% = 99,67%
AB = 1 – log (Δ λ2
λ2) = 1 – log (
0.00050,15 ) = 1−¿(-2,47) = 3,47≈3 AB
Pelaporan:λ2± Δ λ2=(0,15±0,0005 )m
λ2+Δ λ2= (0,15+0,0005 )=0,1505m
λ2+Δ λ2= (0,15+0,0005 )=0,1495m
Δ λ3=| ∂ λ3
∂ X0 X |=|2n|Δs=2
3(0,0005 )=0,0003m
KTPR =Δ λ3
λ3×100 %=
0,00030,12
×100 %=0,25%
Ketelitian = 100%−KTPR= 100% −¿ 0,25% = 99, 75%
AB = 1 – log (Δ λ3
λ3)= 1 – log (
0.00030,12 )= 1−¿(-2,60) = 3,60≈ 4 AB
Pelaporan:λ3± Δ λ3=(0,12±0,0003 )m
λ3+Δ λ3=(0,12+0,0003 )=0,1203m
λ3−Δ λ3=(0,12−0,0003 )m=0,1197m
2. Pada Tali Hijau
Untuk m1 = 0,07 kg
Δ λ1=| ∂ λ1
∂ X0 X |=|2n|Δs=2
1( 0,0005 )=0,001m
KTPR ¿Δ λ1
λ1×100 %=0,001
0,44×100 %=0,23 %
Ketelitian¿100 %−KTPR=100 %−0,23 %=99,77 %
AB = 1 – log (Δ λ1
λ1) = 1 – log (
0.0010,44 ) = 1−¿(-3,64) = 4,64≈5 AB
Pelaporan:λ1± Δ λ1=(0,44±0,001 )m
λ1+Δ λ1= (0,44+0,001 )m=0,441m
λ1−Δ λ1=(0,44−0,001 )m=0,439m
Δ λ2=| ∂ λ2
∂ X0 X |=|2n|Δ s=2
2(0,0005 )=0,0005m
KTPR =Δ λ2
λ2×100 %=
0,00050,24
×100 %=0,21%
Ketelitian = 100%−KTPR= 100% −¿ 0,21% = 99,79%
AB = 1 – log (Δ λ2
λ2) = 1 – log (
0.00050,24 ) = 1−¿(-2,68) = 3,68≈4 AB
Pelaporan:λ2± Δ λ2=(0,24±0,0005 )m
λ2+Δ λ2= (240+0,5 )×10−3m=240,5×10−3m
λ2−Δ λ2=(240−0,5 )×10−3m=2399,5×10−3m
Δ λ3=| ∂ λ3
∂ X0 X |=|2n|Δ s=2
3(0,0005)=0,0003m
KTPR =Δ λ3
λ3×100 %=
0,00030,167
×100 %=0,17 %
Ketelitian = 100%−KTPR= 100% −¿ 0,17% = 99, 83%
AB = 1 – log (Δ λ3
λ3)= 1 – log (
0.00030,167 )= 1−¿(-2,76) = 3,76≈4 AB
Pelaporan:λ3± Δ λ3=(0,167±0,0003 )m
λ3+Δ λ3=(167+0,3 )×10−3m=167,3×10−3m
λ3−Δ λ3=(167−0,3 )×10−3m=166,7×10−3m
Untuk m2 = 0,09 kg
Δ λ1=| ∂ λ1
∂ X0 X |=|2n|Δs=2
1( 0,0005 )=0,001m
KTPR ¿Δ λ1
λ1×100 %=0,001
0,48×100 %=0,18 %
Ketelitian¿100 %−KTPR=100 %−0,18 %=99,82 %
AB = 1 – log (Δ λ1
λ1) = 1 – log (
0.0010,48 ) = 1−¿(-2,73) = 3,73≈4 AB
Pelaporan:λ1± Δ λ1=(0,48±0,001 )m
λ1+Δ λ1= (0,48+0,001 )m=0,481m
λ1−Δ λ1=(0,48−0,001 )m=0,479m
Δ λ2=| ∂ λ2
∂ X0 X |=|2n|Δs=2
2(0,0005 )=0,0005m
KTPR =Δ λ2
λ2×100 %=
0,00050,36
×100%=0,14%
Ketelitian = 100%−KTPR= 100% −¿ 0,14% = 99,86%
AB = 1 – log (Δ λ2
λ2) = 1 – log (
0.00050,36 ) = 1−¿(-2,85) = 3,85≈4 AB
Pelaporan:λ2± Δ λ2=(0,36±0,0005 )m
λ2+Δ λ2= (0,36+0,0005 )m=0,3605m
λ2−Δ λ2=(0,36−0,0005 )m=0,3595m
Δ λ3=| ∂ λ3
∂ X0 X |=|2n|Δs=2
3(0,0005 )=0,0003m
KTPR =Δ λ3
λ3×100 %=
0,00030,186
×100 %=0,16 %
Ketelitian = 100%−KTPR= 100% −¿ 0,16% = 99, 84%
AB = 1 – log (Δ λ3
λ3)= 1 – log (
0.00030,186 )= 1−¿(-2,79) = 3,79≈4 AB
Pelaporan:λ3± Δ λ3=(0,14±0,0003 )m
λ3+Δ λ3=(0,186+0,0003 )m=0,1863m
λ3+Δ λ3=(0,186−0,0003 )m=0,1857m
Untuk m3 = 0,10 kg
Δ λ1=| ∂ λ1
∂ X0 X |=|2n|Δs=2
1( 0,0005 )=0,001m
KTPR ¿Δ λ1
λ1×100 %=0,001
0,44×100 %=0,23 %
Ketelitian¿100 %−KTPR=100 %−0,23 %=99,77 %
AB = 1 – log (Δ λ1
λ1) = 1 – log (
0.0010,44 ) = 1−¿(-3,64) = 4,64≈5 AB
Pelaporan:λ1± Δ λ1=(0,44±0,001 )m
λ1+Δ λ1= (0,44+0,001 )m=0,441m
λ1−Δ λ1=(0,44−0,001 )m=0,439m
Δ λ2=| ∂ λ2
∂ X0 X |=|2n|Δs=2
2(0,0005 )=0,0005m
KTPR =Δ λ2
λ2×100 %=
0,00050,26
×100 %=0,19%
Ketelitian = 100%−KTPR= 100% −¿ 0,19% = 99,81%
AB = 1 – log (Δ λ2
λ2) = 1 – log (
0.00050,26 ) = 1−¿(-2,71) = 3,71≈4 AB
Pelaporan:λ2± Δ λ2=(0,26±0,0005 )m
λ2+Δ λ2= (0,26+0,0005 )=0,2605m
λ2+Δ λ2= (0,26+0,0005 )=0,2595m
Δ λ3=| ∂ λ3
∂ X0 X |=|2n|Δs=2
3(0,0005 )=0,0003m
KTPR =Δ λ3
λ3×100 %=
0,00030,20
×100 %=0,15%
Ketelitian = 100%−KTPR= 100% −¿ 0,15% = 99, 85%
AB = 1 – log (Δ λ3
λ3)= 1 – log (
0.00030,20 )= 1−¿(-2,82) = 3,83≈ 4 AB
Pelaporan:λ3± Δ λ3=(0,20±0,0003 )m
λ3+Δ λ3=(0,20+0,0003 )=0,2003m
λ3−Δ λ3=(0,20−0,0003 )m=0,1997m
3. Pada Tali Kuning
Untuk m1 = 0,02 kg
Δ λ1=| ∂ λ1
∂ X0 X |=|2n|Δs=2
1( 0,0005 )=0,001m
KTPR ¿Δ λ1
λ1×100 %=0,001
0,56×100 %=0,18 %
Ketelitian¿100 %−KTPR=100 %−0,18 %=99,82 %
AB = 1 – log (Δ λ1
λ1) = 1 – log (
0.0010,56 ) = 1−¿(-2,74) = 4,74≈4 AB
Pelaporan:λ1± Δ λ1=(0,56±0,001 )m
λ1+Δ λ1= (0,56+0,001 )m=0,561m
λ1−Δ λ1=(0,56−0,001 )m=0,559m
Δ λ2=| ∂ λ2
∂ X0 X |=|2n|Δ s=2
2(0,0005 )=0,0005m
KTPR =Δ λ2
λ2×100 %=
0,00050,23
×100 %=0,21%
Ketelitian = 100%−KTPR= 100% −¿ 0,21% = 99,79%
AB = 1 – log (Δ λ2
λ2) = 1 – log (
0.00050,23 ) = 1−¿(-2,68) = 3,68≈4 AB
Pelaporan:λ2± Δ λ2=(0,23±0,0005 )m
λ2+Δ λ2= (230+0,5 )×10−3m=230,5×10−3m
λ2−Δ λ2=(230−0,5 )×10−3m=229,5×10−3m
Δ λ3=| ∂ λ3
∂ X0 X |=|2n|Δ s=2
3(0,0005)=0,0003m
KTPR =Δ λ3
λ3×100 %=
0,00030,18
×100 %=0,16 %
Ketelitian = 100%−KTPR= 100% −¿ 0,16% = 99, 84%
AB = 1 – log (Δ λ3
λ3)= 1 – log (
0.00030,18 )= 1−¿(2,77) = 3,77≈4 AB
Pelaporan:λ3± Δ λ3=(0,18±0,0003 )m
λ3+Δ λ3=(18+0,3 )×10−3m=18,3×10−3m
λ3−Δ λ3=(18−0,3 )×10−3m=17,7×10−3m
Untuk m2 = 0,03 kg
Δ λ1=| ∂ λ1
∂ X0 X |=|2n|Δs=2
1( 0,0005 )=0,001m
KTPR ¿Δ λ1
λ1×100 %=0,001
0,36×100 %=0,27 %
Ketelitian¿100 %−KTPR=100 %−0,27 %=99,73 %
AB = 1 – log (Δ λ1
λ1) = 1 – log (
0.0010,36 ) = 1−¿(-2,55) = 3,55≈4 AB
Pelaporan:λ1± Δ λ1=(0,36±0,001 )m
λ1+Δ λ1= (0,36+0,001 )m=0,361m
λ1−Δ λ1=(0,36−0,001 )m=0,359m
Δ λ2=| ∂ λ2
∂ X0 X |=|2n|Δs=2
2(0,0005 )=0,0005m
KTPR =Δ λ2
λ2×100 %=
0,00050,36
×100%=0,14%
Ketelitian = 100%−KTPR= 100% −¿ 0,14% = 99,86%
AB = 1 – log (Δ λ2
λ2) = 1 – log (
0.00050,36 ) = 1−¿(-2,85) = 3,85≈4 AB
Pelaporan:λ2± Δ λ2=(0,36±0,0005 )m
λ2+Δ λ2= (0,36+0,0005 )m=0,3605m
λ2−Δ λ2=(0,36−0,0005 )m=0,3595m
Δ λ3=| ∂ λ3
∂ X0 X |=|2n|Δs=2
3(0,0005 )=0,0003m
KTPR =Δ λ3
λ3×100 %=
0,00030,226
×100 %=0,13%
Ketelitian = 100%−KTPR= 100% −¿ 0,13% = 99, 87%
AB = 1 – log (Δ λ3
λ3)= 1 – log (
0.00030,226 )= 1−¿(-2,87) = 3,87≈4 AB
Pelaporan:λ3± Δ λ3=(0,266±0,0003 )m
λ3+Δ λ3=(0,226+0,0003 )m=0,2263m
λ3+Δ λ3=(0,226−0,0003 )m=0,2257m
Untuk m3 = 0,04 kg
Δ λ1=| ∂ λ1
∂ X0 X |=|2n|Δs=2
1( 0,0005 )=0,001m
KTPR ¿Δ λ1
λ1×100 %=0,001
0,14×100 %=0,71%
Ketelitian¿100 %−KTPR=100 %−0,71 %=99,29 %
AB = 1 – log (Δ λ1
λ1) = 1 – log (
0.0010,14 ) = 1−¿(-2,14) = 3,14≈3 AB
Pelaporan:λ1± Δ λ1=(0,14±0,001 )m
λ1+Δ λ1= (0,14+0,001 )m=0,141m
λ1−Δ λ1=(0,14−0,001 )m=0,139m
Δ λ2=| ∂ λ2
∂ X0 X |=|2n|Δs=2
2(0,0005 )=0,0005m
KTPR =Δ λ2
λ2×100 %=
0,00050,36
×100%=0,14%
Ketelitian = 100%−KTPR= 100% −¿ 0,14% = 99,86%
AB = 1 – log (Δ λ2
λ2) = 1 – log (
0.00050,36 ) = 1−¿(-2,85) = 3,85≈4 AB
Pelaporan:λ2± Δ λ2=(0,36±0,0005 )m
λ2+Δ λ2= (0,36+0,0005 )m=0,3605m
λ2−Δ λ2=(0,36−0,0005 )m=0,3595m
Δ λ3=| ∂ λ3
∂ X0 X |=|2n|Δs=2
3(0,0005 )=0,0003m
KTPR =Δ λ3
λ3×100 %=
0,00030,16
×100 %=0,19%
Ketelitian = 100%−KTPR= 100% −¿ 0,19% = 99, 81%
AB = 1 – log (Δ λ3
λ3)= 1 – log (
0.00030,16 )= 1−¿(-2,72) = 3,72≈ 4 AB
Pelaporan:λ3± Δ λ3=(0,16±0,0003 )m
λ3+Δ λ3=(0,16+0,0003 )=0,1603m
λ3−Δ λ3=(0,16−0,0003 )m=0,1597m
B. Menghitung Gaya TeganganTali
∆ FT=|∂FT
∂m ||∆m|=|g||∆ m|
Dimana :
∆ m=12NST Neracadigital=5×10−6 Kg
1) Pada Tali Putih
∆ FT 1=|∂F∂m||Δm|=g . Δm=(9,8)(5×10−6)=4,9×10−5N
KTPR=∆ FT 1
FT 1×100 %=4,9×10−5
0,686×100 %=0,23 %
Ketelitian=100 %−KTPR=100 %−0,01%=99,77 %
AB=1− log∆ FT 1
FT 1=1−¿ log 4,9×10−5
0,686¿5≈5 AB ¿
Pelaporan=(FT 1±∆ FT 1 )= (0,49±0,00005 ) N
(FT 1+∆ FT 1 )= (49000+5 ) 10−5 N=49005×10−5 N
(FT 1−∆FT 1 )=(49000−5 ) 10−5N=48995×10−5N
∆ FT 2=|∂F∂m||Δm|=g . Δm=(9,8)(5×10−6)=4,9×10−5N
KTPR=∆ FT 2
FT 2×100 %=4,9×10−5
0,882×100 %=0,005 %
Ketelitian=100 %−KTPR=100 %−0,005 %=99,995 %
AB=1− log∆ FT 2
FT 2=1−¿ log 4,9×10−5
0,882¿4,25≈ 4 AB ¿
Pelaporan=(FT 2±∆ FT 2 )= (0,882±0,00005 ) N
(FT 2+∆ FT 2 )= (88200+5 )×10−5 N=88205×10−5N
(FT 2−∆FT 2 )=(88200−5 )×10−5 N=88195×10−5N
∆ FT 3=|∂F∂m||Δm||9,8|=g . Δm=(9,8)(5×10−6)=4,9×10−5N
KTPR=∆ FT 3
FT 3×100 %=4,9×10−5
1,078=0,004 %
Ketelitian=100 %−KTPR=100 %−0,004 %=99,996 %
AB=1−log∆ FT 3
FT 3=1−¿ log 4,9×10−5
1,0784,34≈4 AB ¿
Pelaporan=(FT 3±∆ FT 3 )=(1,078±0,00005 ) N
(FT 3+∆ FT 3 )=(107800+5 )×10−5 N=107805×10−5N
(FT 3−∆ FT 3 )=(107800−5 )×10−5N=107795×10−5N
2) Pada Tali Hijau
∆ FT 1=|∂F∂m||Δm|=g . Δm=(9,8)(5×10−6)=4,9×10−5N
KTPR=∆ FT 1
FT 1×100 %=4,9×10−5
0,686×100 %=0,23 %
Ketelitian=100 %−KTPR=100 %−0,01%=99,77 %
AB=1−log∆ FT 1
FT 1=1−¿ log 4,9×10−5
0,686¿5≈5 AB ¿
Pelaporan=(FT 1±∆ FT 1 )= (0,49±0,00005 ) N
(FT 1+∆ FT 1 )= (49000+5 ) 10−5 N=49005×10−5 N
(FT 1−∆FT 1 )=(49000−5 ) 10−5N=48995×10−5N
∆ FT 2=|∂F∂m||Δm|=g . Δm=(9,8)(5×10−6)=4,9×10−5N
KTPR=∆ FT 2
FT 2×100 %=4,9×10−5
0,882×100 %=0,005 %
Ketelitian=100 %−KTPR=100 %−0,005 %=99,995 %
AB=1− log∆ FT 2
FT 2=1−¿ log 4,9×10−5
0,882¿4,25≈ 4 AB ¿
Pelaporan=(FT 2±∆ FT 2 )= (0,882±0,00005 ) N
(FT 2+∆ FT 2 )= (88200+5 )×10−5 N=88205×10−5N
(FT 2−∆FT 2 )=(88200−5 )×10−5 N=88195×10−5N
∆ FT 3=|∂F∂m||Δm||9,8|=g . Δm=(9,8)(5×10−6)=4,9×10−5N
KTPR=∆ FT 3
FT 3×100 %= 4,9×10−5
0,98=0,005 %
Ketelitian=100 %−KTPR=100 %−0,005 %=99,995 %
AB=1− log∆ FT 3
FT 3=1−¿ log 4,9×10−5
0,98=4,30≈ 4 AB ¿
Pelaporan=(FT 3±∆ FT 3 )=(0,98±0,00005 )N
(FT 3+∆ FT 3 )=(980000+5 )×10−5 N=980005×10−5 N
(FT 3−∆ FT 3 )=(980000−5 )×10−5 N=979995×10−5N
3) Pada Tali Kuning
∆ FT 1=|∂F∂m||Δm|=g . Δm=(9,8)(5×10−6)=4,9×10−5N
KTPR=∆ FT 1
FT 1×100 %=4,9×10−5
0,196×100 %=0,02 %
Ketelitian=100 %−KTPR=100 %−0,02%=99,92 %
AB=1−log∆ FT 1
FT 1=1−¿ log 4,9×10−5
0,196¿4,60≈5 AB¿
Pelaporan=(FT 1±∆ FT 1 )= (0,196±0,00005 ) N
(FT 1+∆ FT 1 )= (19600+5 )10−5N=19605×10−5 N
(FT 1−∆FT 1 )=(19600−5 )10−5N=19595×10−5N
∆ FT 2=|∂F∂m||Δm||9,8|=g . Δm=(9,8)(5×10−6)=4,9×10−5 N
KTPR=∆ FT 2
FT 2×100%=4,9×10−5
0,294×100 %=0,016 %
Ketelitian=100 %−KTPR=100 %−0,016 %=99,984 %
AB=1−log∆ FT 2
FT 2=1−¿ log 4,9×10−5
0,294¿4,77≈5 AB¿
Pelaporan=(FT 2±∆ FT 2 )= (0,98±0,00005 ) N
(FT 2+∆ FT 2 )= (294000+5 )×10−5 N=294005×10−5N
(FT 2−∆FT 2 )=(294000−5 )×10−5N=293995×10−5N
∆ FT 3=|∂F∂m||Δm||9,8|=g . Δm=(9,8)(5×10−6)=4,9×10−5N
KTPR=∆ FT 3
FT 3×100 %= 4,9×10−5
0,392=0,003 %
Ketelitian=100 %−KTPR=100 %−0,003 %=99,997 %
AB=1− log∆ FT 3
FT 3=1−¿ log 0,392=5,48≈5 AB ¿
Pelaporan=(FT 3±∆ FT 3 )=(0,392±0,00005 ) N
(FT 3+∆ FT 3 )=(392000+5 )×10−5 N=392005×10−5N
(FT 3−∆ FT 3 )=(392000−5 )×10−5 N=391995×10−5N
C. Menghitung Laju Rambat Gelombang Secara Eksperimen
∆ v=|∂v∂ f ||∆ λ|=|f||∆ λ|
Dimana :
f=50Hz
1) Tali Putih
Untuk m1 =0,07 kg
∆ v1=|50||1×10−3|=0,05 ms
KTPR=∆v1
v1×100 %=0,05
10×100 %=0,5 %
Ketelitian=100 %−KTPR=100 %−0,5 %=99,5 %
AB=1− log∆ v1
v1=1−log 0,05
10¿3,30≈3 AB
Pelaporan=(v1±∆v1 )=(10±0,05 ) ms
(v1+∆v1 )= (10+0,05 ) ms=10,05 m
s
(v1−∆ v1)=(10−0,05 ) ms=9,95 m
s
∆ v2=|50||5×10−4|=0,025 ms
KTPR=∆v2
v2×100 %=0,025
5,5×100 %=0,08 %
Ketelitian=100 %−KTPR=100 %−0,08 %=99,92%
AB=1− log∆ v2
v2=1−log 0,05
5,5¿4,08≈ 4 AB
Pelaporan=(v2±∆v2 )=(5,5±0,025 ) N
(v2+∆v2 )=(550+2,5 )×10−2 ms=552,5×10−2 m
s
(v2−∆ v2)=(550−2,5 )×10−1 ms=547,5×10−2 m
s
∆ v3=|50||3×10−4|=0,017 ms
KTPR=∆v3
v3×100%=0,017
4,3×100 %=0,39%
Ketelitian=100 %−KTPR=100 %−0,39 %=99,61%
AB=1−log∆ v3
v3=1−log 0,017
4,3¿3,40≈3 AB
Pelaporan=(v3±∆v3 )=(4,3±0,017 ) ms
(v3+∆v3 )=(430+1,7 )×10−2 ms=437,1×10−2 m
s
(v3−∆ v3 )=( 430−1,7 )×10−2 ms=428,3×10−2 m
s
Untuk m2 = 0,09 kg
∆ v1=|50||1×10−3|=0,05 ms
KTPR=∆v1
v1×100 %=0,05
27×100 %=0,18 %
Ketelitian=100 %−KTPR=100 %−0,18 %=99,82%
AB=1− log∆ v1
v1=1−log 0,05
27¿3,73≈4 AB
Pelaporan=(v1±∆v1 )=(27±0,05 )N
(v1+∆v1 )= (27+0,05 ) ms=27,05 m
s
(v1−∆ v1)=(27−0,05 ) ms=26,95 m
s
∆ v2=|50||5×10−4|=0,025 ms
KTPR=∆v2
v2×100 %=0,025
14×100 %=0,17 %
Ketelitian=100 %−KTPR=100 %−0,17 %=99,83 %
AB=1−log∆ v2
v2=1−log 0,025
14¿3,74≈4 AB
Pelaporan=(v2±∆v2 )=(14±0,025 ) ms
(v2+∆v2 )=(14+0,025 ) ms=14,02 m
s
(v2−∆ v2)=(14−0,025 ) ms=13,97 m
s
∆ v3=|50||3×10−4|=0,017 ms
KTPR=∆v3
v3×100 %=0,017
7×100 %=0,24 %
Ketelitian=100 %−KTPR=100 %−0,24 %=99,76 %
AB=1− log∆ v3
v3=1−log 0,017
7¿3,61≈4 AB
Pelaporan=(v3±∆v3 )=(7±0,017 ) ms
(v3+∆v3 )=(7+0,017 ) ms=7,017m
s
(v3−∆ v3 )=(7−0,017 ) ms=765,3 m
s
Untuk m3 = 0,11 kg
∆ v1=|50||1×10−3|=0,05 ms
KTPR=∆v1
v1×100 %=0,05
11×100 %=0,45 %
Ketelitian=100 %−KTPR=100 %−0,45%=99,55 %
AB=1− log∆ v1
v1=1−log 0,05
11¿3,34≈4 AB
Pelaporan=(v1±∆v1 )=(11±0,05 )N
(v1+∆v1 )= (11+0,05 ) ms=11,05 m
s
(v1−∆ v1)=(11−0,05 ) ms=10,95 m
s
∆ v2=|50||5×10−4|=0,025 ms
KTPR=∆v2
v2×100 %=0,025
7,5×100 %=0,33 %
Ketelitian=100 %−KTPR=100 %−0,33 %=99,67 %
AB=1−log∆ v2
v2=1−log 0,025
7,5¿3,48≈3 AB
Pelaporan=(v2±∆v2 )=(7,5±0,025 ) ms
(v2+∆v2 )=(7,5+0,025 ) ms=7,52m
s
(v2−∆ v2)=(7,5−0,025 ) ms=7,47 m
s
∆ v3=|50||3×10−4|=0,017 ms
KTPR=∆v3
v3×100 %=0,017
6×100 %=0,28 %
Ketelitian=100 %−KTPR=100 %−0,28 %=99,72%
AB=1− log∆ v3
v3=1−log 0,017
6¿3,55≈4 AB
Pelaporan=(v3±∆v3 )=(6±0,017 ) ms
(v3+∆v3 )=(6+0,017 ) ms=6,017m
s
(v3−∆ v3 )=(6−0,017 ) ms=5,983 m
s
2) Pada Tali Hijau
Untuk m1 = 0,07 kg
∆ v1=|50||1×10−3|=0,05 ms
KTPR=∆v1
v1×100 %=0,05
22×100 %=0,22 %
Ketelitian=100 %−KTPR=100 %−0,22 %=99,78 %
AB=1− log∆ v1
v1=1−log 0,05
22¿3,64≈4 AB
Pelaporan=(v1±∆v1 )=(22±0,05 ) ms
(v1+∆v1 )= (22+0,05 ) ms=22,05m
s
(v1−∆ v1)=(22−0,05 ) ms=21,95m
s
∆ v2=|50||5×10−4|=0,025 ms
KTPR=∆v2
v2×100 %=0,025
12×100 %=0,21 %
Ketelitian=100 %−KTPR=100 %−0,21 %=99,79 %
AB=1− log∆ v2
v2=1−log 0,025
12¿3,68≈4 AB
Pelaporan=(v2±∆v2 )=(12±0,025 ) ms
(v2+∆v2 )=(12+0,025 ) ms=12,02m
s
(v2−∆ v2)=(12−0,025 ) ms=11,97 m
s
∆ v3=|50||3×10−4|=0,017 ms
KTPR=∆v3
v3×100 %=0,017
8,35×100 %=0,20 %
Ketelitian=100 %−KTPR=100 %−0,20 %=99,80 %
AB=1− log∆ v3
v3=1−log 0,017
8,35¿3,69≈4 AB
Pelaporan=(v3±∆v3 )=(8,35±0,017 ) ms
(v3+∆v3 )=(8,35+0,017 ) ms=8,367m
s
(v3−∆ v3 )=(8,35−0,017 ) ms=8,333m
s
Untuk m2= 0,09 kg
∆ v1=|50||1×10−3|=0,05 ms
KTPR=∆v1
v1×100 %=0,05
24×100 %=0,21 %
Ketelitian=100 %−KTPR=100 %−0,21 %=99,79 %
AB=1− log∆ v1
v1=1−log 0,05
24¿3,68≈4 AB
Pelaporan=(v1±∆v1 )=(24±0,05 ) ms
(v1+∆v1 )= (24+0,05 ) ms=24,05m
s
(v1−∆ v1)=(24−0,05 ) ms=23,95 m
s
∆ v2=|50||5×10−4|=0,025 ms
KTPR=∆v2
v2×100 %=0,025
18×100 %=0,14 %
Ketelitian=100 %−KTPR=100 %−0,14 %=99,86 %
AB=1−log∆ v2
v2=1−log 0,025
18¿3,85≈4 AB
Pelaporan=(v2±∆v2 )=(18±0,025 ) ms
(v2+∆v2 )=(18+0,025 ) ms=18,02m
s
(v2−∆ v2)=(18−0,025 ) ms=17,97 m
s
∆ v3=|50||3×10−4|=0,017 ms
KTPR=∆v3
v3×100 %=0,017
9,3×100 %=0,18 %
Ketelitian=100 %−KTPR=100 %−0,18 %=99,82%
AB=1−log∆ v3
v3=1−log 0,017
9,3¿3,73≈4 AB
Pelaporan=(v3±∆v3 )=(9,3±0,017 ) ms
(v3+∆v3 )=(9,3+0,017 ) ms=9,317 m
s
(v3−∆ v3 )=(9,3−0,017 ) ms=9,283m
s
Untuk m3= 0,10 kg
∆ v1=|50||1×10−3|=0,05 ms
KTPR=∆v1
v1×100%=0,05
11×100 %=0,45%
Ketelitian=100 %−KTPR=100 %−0,45%=99,55 %
AB=1−log∆ v1
v1=1−log 0,05
11¿3,34≈3 AB
Pelaporan=(v1±∆v1 )=(11±0,05 ) ms
(v1+∆v1 )= (11+0,05 ) ms=11,1 m
s
(v1−∆ v1)=(11−0,05 ) ms=10,9 m
s
∆ v2=|50||5×10−4|=0,025 ms
KTPR=∆v2
v2×100%=0,025
13×100 %=0,19%
Ketelitian=100 %−KTPR=100 %−0,19 %=99,81%
AB=1−log∆ v2
v2=1−log 0,025
13¿3,71≈4 AB
Pelaporan=(v2±∆v2 )=(13±0,025 ) ms
(v2+∆v2 )=(13+0,025 ) ms=13,02m
s
(v2−∆ v2)=(13−0,025 ) ms=12,97 m
s
∆ v3=|50||3×10−4|=0,017 ms
KTPR=∆v3
v3×100 %=0,017
10×100 %=0,17 %
Ketelitian=100 %−KTPR=100 %−0,17 %=99,83 %
AB=1− log∆ v3
v3=1−log 0,017
10¿3,76≈ 4 AB
Pelaporan=(v3±∆v3 )=(10±0,017 ) ms
(v3+∆v3 )=(10+0,017 ) ms=10,02 m
s
(v3−∆ v3 )=(10−0,017 ) ms=9,983m
s
3) Pada Tali Kuning
Untuk m1 = 0,02 kg
∆ v1=|50||1×10−3|=0,05 ms
KTPR=∆v1
v1×100 %=0,05
28×100 %=0,17 %
Ketelitian=100 %−KTPR=100 %−0,17 %=99,83 %
AB=1− log∆ v1
v1=1−log 0,05
28¿3,75≈4 AB
Pelaporan=(v1±∆v1 )=(28±0,05 ) ms
(v1+∆v1 )= (28+0,05 ) ms=28,05 m
s
(v1−∆ v1)=(28−0,05 ) ms=27,95m
s
∆ v2=|50||5×10−4|=0,025 ms
KTPR=∆v2
v2×100 %=0,025
11,5×100 %=0,22 %
Ketelitian=100 %−KTPR=100 %−0,22 %=99,78 %
AB=1− log∆ v2
v2=1−log 0,025
11,5¿3,66≈4 AB
Pelaporan=(v2±∆v2 )=(11,5±0,025 ) ms
(v2+∆v2 )=(11,5+0,025 ) ms=11,52 m
s
(v2−∆ v2)=(11,5−0,025 ) ms=11,47 m
s
∆ v3=|50||3×10−4|=0,017 ms
KTPR=∆v3
v3×100 %=0,017
9×100 %=0,19 %
Ketelitian=100 %−KTPR=100 %−0,19 %=99,81%
AB=1−log∆ v3
v3=1−log 0,017
9¿3,72≈4 AB
Pelaporan=(v3±∆v3 )=(9±0,017 ) ms
(v3+∆v3 )=(9+0,017 ) ms=9,017 m
s
(v3−∆ v3 )=(9−0,017 ) ms=8,983m
s
Untuk m2 = 0,03 kg
∆ v1=|50||1×10−3|=0,05 ms
KTPR=∆v1
v1×100 %=0,05
18×100 %=0,28 %
Ketelitian=100 %−KTPR=100 %−0,28 %=99,72%
AB=1−log∆ v1
v1=1−log 0,05
18¿3,55≈4 AB
Pelaporan=(v1±∆v1 )=(18±0,05 ) ms
(v1+∆v1 )= (18+0,05 ) ms=18,05 m
s
(v1−∆ v1)=(18−0,05 ) ms=17,95m
s
∆ v2=|50||5×10−4|=0,025 ms
KTPR=∆v2
v2×100%=0,025
18×100 %=0,14%
Ketelitian=100 %−KTPR=100 %−0,14 %=99,86 %
AB=1−log∆ v2
v2=1−log 0,025
18¿3,85≈4 AB
Pelaporan=(v2±∆v2 )=(18±0,025 ) ms
(v2+∆v2 )=(18+0,025 ) ms=18,02m
s
(v2−∆ v2)=(18−0,025 ) ms=17,97 m
s
∆ v3=|50||3×10−4|=0,017 ms
KTPR=∆v3
v3×100 %=0,017
11,3×100 %=0,15 %
Ketelitian=100 %−KTPR=100 %−0,19 %=99,85 %
AB=1−log∆ v3
v3=1−log 0,017
11,3¿3,82≈4 AB
Pelaporan=(v3±∆v3 )=(11,3±0,017 ) ms
(v3+∆v3 )=(11,3+0,017 ) ms=11,04 m
s
(v3−∆ v3 )=(11,3−0,017 ) ms=11,28 m
s
Untuk m3 = 0,04 kg
∆ v1=|50||1×10−3|=0,05 ms
KTPR=∆v1
v1×100 %=0,05
14×100 %=0,35 %
Ketelitian=100 %−KTPR=100 %−0,35%=99,65 %
AB=1− log∆ v1
v1=1−log 0,05
14¿3,44≈3 AB
Pelaporan=(v1±∆v1 )=(14±0,05 ) ms
(v1+∆v1 )= (14+0,05 ) ms=14,1 m
s
(v1−∆ v1)=(14−0,05 ) ms=13,9 m
s
∆ v2=|50||5×10−4|=0,025 ms
KTPR=∆v2
v2×100 %=0,025
19×100 %=0,13 %
Ketelitian=100 %−KTPR=100 %−0,13 %=99,87 %
AB=1− log∆ v2
v2=1−log 0,025
19¿3,88≈4 AB
Pelaporan=(v2±∆v2 )=(19±0,025 ) ms
(v2+∆v2 )=(19+0,025 ) ms=19,02m
s
(v2−∆ v2)=(19−0,025 ) ms=18,97 m
s
∆ v3=|50||3×10−4|=0,017 ms
KTPR=∆v3
v3×100 %=0,017
8×100 %=0,21 %
Ketelitian=100 %−KTPR=100 %−0,21 %=99,79 %
AB=1− log∆ v3
v3=1−log 0,017
8¿3,67≈ 4 AB
Pelaporan=(v3±∆v3 )=(8±0,017 ) ms
(v3+∆v3 )=(8+0,017 ) ms=8,017m
s
(v3−∆ v3 )=(8−0,017 ) ms=7,983 m
s