Trong một dao động bất kì, mối liên hệ giữa tần số góc ω \omega ω , chu kì T , tần số f bằng công thức nào?

ω = 2 π T = 2 π . f \omega=\frac{2\pi}{T}=2\pi.f ω = T 2 π = 2 π . f

ω = 2 π f = 2 π . T \omega=\frac{2\pi}{f}=2\pi.T ω = f 2 π = 2 π . T

T = 2 π ω = 2 π . f T=\frac{2\pi}{\omega}=2\pi.f T = ω 2 π = 2 π . f

T = 2 π f = 2 π . ω T=\frac{2\pi}{f}=2\pi.\omega T = f 2 π = 2 π . ω

Công thức tính chu kì của con lắc lò xo

T = 2 π m k T=\ 2\pi\sqrt{\frac{m}{k}} T = 2 π k m <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

T = m k T=\ \sqrt{\frac{m}{k}} T = k m <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

T = k m T=\ \sqrt{\frac{k}{m}} T = m k <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

T = 2 π k m T=\ 2\pi\sqrt{\frac{k}{m}} T = 2 π m k <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

Một con lắc lò xo có độ cứng k, vật nặng khối lượng m. Tần số dao động của con lắc được xác định bởi hệ thức

f = 1 2 π k m f\ =\frac{1}{2\pi}\sqrt{\frac{k}{m}} f = 2 π 1 m k <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

f = 1 2 π m k f\ =\frac{1}{2\pi}\sqrt{\frac{m}{k}} f = 2 π 1 k m <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

f = 2 π m k f\ =2\pi\sqrt{\frac{m}{k}} f = 2 π k m <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

f = 2 π k m f\ =2\pi\sqrt{\frac{k}{m}} f = 2 π m k <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

Công thức tính tần số góc của con lắc lò xo

ω = k m \omega=\sqrt{\frac{k}{m}} ω = m k <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

ω = m k \omega=\sqrt{\frac{m}{k}} ω = k m <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

ω = 2 π . k m \omega=2\pi.\sqrt{\frac{k}{m}} ω = 2 π . m k <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

ω = 2 π . m k \omega=2\pi.\sqrt{\frac{m}{k}}\ ω = 2 π . k m <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

Công thức tính giá trị lực đàn hồi của con lắc lò xo

F = k ∣ x ∣ F=k\left|x\right| F = k ∣ x ∣

F = k . ω F=k.\omega F = k . ω

Công thức tính chu kì của con lắc đơn

T = 2 π l g T=\ 2\pi\sqrt{\frac{l}{g}} T = 2 π g l <path d="M473,2793c339.3,-1799.3,509.3,-2700,510,-2702 c3.3,-7.3,9.3,-11,18,-11H400000v40H1017.7s-90.5,478,-276.2,1466c-185.7,988, -279.5,1483,-281.5,1485c-2,6,-10,9,-24,9c-8,0,-12,-0.7,-12,-2c0,-1.3,-5.3,-32, -16,-92c-50.7,-293.3,-119.7,-693.3,-207,-1200c0,-1.3,-5.3,8.7,-16,30c-10.7, 21.3,-21.3,42.7,-32,64s-16,33,-16,33s-26,-26,-26,-26s76,-153,76,-153s77,-151, 77,-151c0.7,0.7,35.7,202,105,604c67.3,400.7,102,602.7,104,606z M1001 80H400000v40H1017z">

T = 2 l g T=\ 2\sqrt{\frac{l}{g}} T = 2 g l <path d="M473,2793c339.3,-1799.3,509.3,-2700,510,-2702 c3.3,-7.3,9.3,-11,18,-11H400000v40H1017.7s-90.5,478,-276.2,1466c-185.7,988, -279.5,1483,-281.5,1485c-2,6,-10,9,-24,9c-8,0,-12,-0.7,-12,-2c0,-1.3,-5.3,-32, -16,-92c-50.7,-293.3,-119.7,-693.3,-207,-1200c0,-1.3,-5.3,8.7,-16,30c-10.7, 21.3,-21.3,42.7,-32,64s-16,33,-16,33s-26,-26,-26,-26s76,-153,76,-153s77,-151, 77,-151c0.7,0.7,35.7,202,105,604c67.3,400.7,102,602.7,104,606z M1001 80H400000v40H1017z">

T = 2 g l T=\ 2\sqrt{\frac{g}{l}} T = 2 l g <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

T = 2 π g l T=\ 2\pi\sqrt{\frac{g}{l}} T = 2 π l g <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

Công thức tính tần số góc của con lắc đơn

ω = g l \omega=\ \sqrt{\frac{g}{l}} ω = l g <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

ω = l g \omega=\sqrt{\frac{l}{g}} ω = g l <path d="M473,2793c339.3,-1799.3,509.3,-2700,510,-2702 c3.3,-7.3,9.3,-11,18,-11H400000v40H1017.7s-90.5,478,-276.2,1466c-185.7,988, -279.5,1483,-281.5,1485c-2,6,-10,9,-24,9c-8,0,-12,-0.7,-12,-2c0,-1.3,-5.3,-32, -16,-92c-50.7,-293.3,-119.7,-693.3,-207,-1200c0,-1.3,-5.3,8.7,-16,30c-10.7, 21.3,-21.3,42.7,-32,64s-16,33,-16,33s-26,-26,-26,-26s76,-153,76,-153s77,-151, 77,-151c0.7,0.7,35.7,202,105,604c67.3,400.7,102,602.7,104,606z M1001 80H400000v40H1017z">

ω = 2 π . g l \omega=2\pi.\sqrt{\frac{g}{l}} ω = 2 π . l g <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

ω = 2 π . l g \omega=2\pi.\sqrt{\frac{l}{g}}\ ω = 2 π . g l <path d="M473,2793c339.3,-1799.3,509.3,-2700,510,-2702 c3.3,-7.3,9.3,-11,18,-11H400000v40H1017.7s-90.5,478,-276.2,1466c-185.7,988, -279.5,1483,-281.5,1485c-2,6,-10,9,-24,9c-8,0,-12,-0.7,-12,-2c0,-1.3,-5.3,-32, -16,-92c-50.7,-293.3,-119.7,-693.3,-207,-1200c0,-1.3,-5.3,8.7,-16,30c-10.7, 21.3,-21.3,42.7,-32,64s-16,33,-16,33s-26,-26,-26,-26s76,-153,76,-153s77,-151, 77,-151c0.7,0.7,35.7,202,105,604c67.3,400.7,102,602.7,104,606z M1001 80H400000v40H1017z">

Công thức tính tần số của con lắc đơn

f = 1 2 π g l f=\frac{1}{2\pi}\sqrt{\frac{g}{l}} f = 2 π 1 l g <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

f = 2 π g l f=2\pi\sqrt{\frac{g}{l}} f = 2 π l g <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

f = 1 2 π l g f=\frac{1}{2\pi}\sqrt{\frac{l}{g}} f = 2 π 1 g l <path d="M473,2793c339.3,-1799.3,509.3,-2700,510,-2702 c3.3,-7.3,9.3,-11,18,-11H400000v40H1017.7s-90.5,478,-276.2,1466c-185.7,988, -279.5,1483,-281.5,1485c-2,6,-10,9,-24,9c-8,0,-12,-0.7,-12,-2c0,-1.3,-5.3,-32, -16,-92c-50.7,-293.3,-119.7,-693.3,-207,-1200c0,-1.3,-5.3,8.7,-16,30c-10.7, 21.3,-21.3,42.7,-32,64s-16,33,-16,33s-26,-26,-26,-26s76,-153,76,-153s77,-151, 77,-151c0.7,0.7,35.7,202,105,604c67.3,400.7,102,602.7,104,606z M1001 80H400000v40H1017z">

f = 2 π l g f=2\pi\sqrt{\frac{l}{g}} f = 2 π g l <path d="M473,2793c339.3,-1799.3,509.3,-2700,510,-2702 c3.3,-7.3,9.3,-11,18,-11H400000v40H1017.7s-90.5,478,-276.2,1466c-185.7,988, -279.5,1483,-281.5,1485c-2,6,-10,9,-24,9c-8,0,-12,-0.7,-12,-2c0,-1.3,-5.3,-32, -16,-92c-50.7,-293.3,-119.7,-693.3,-207,-1200c0,-1.3,-5.3,8.7,-16,30c-10.7, 21.3,-21.3,42.7,-32,64s-16,33,-16,33s-26,-26,-26,-26s76,-153,76,-153s77,-151, 77,-151c0.7,0.7,35.7,202,105,604c67.3,400.7,102,602.7,104,606z M1001 80H400000v40H1017z">

Biên độ tổng hợp A của 2 dao động điều hòa có giá trị thuộc công thức nào?

∣ A 1 − A 2 ∣ ≤ A ≤ A 1 + A 2 \left|A_1-A_2\right|\le A\le A_1+A_2 ∣ A 1 − A 2 ∣ ≤ A ≤ A 1 + A 2

A 1 ≤ A ≤ A 2 A_1\le A\le A_2 A 1 ≤ A ≤ A 2

A ≥ A 1 + A 2 A\ge A_1+A_2 A ≥ A 1 + A 2

A ≤ A 1 + A 2 A\le A_1+A_2 A ≤ A 1 + A 2

Công thức liên hệ vận tốc v v v , bước sóng λ \lambda λ , tần số f f f

v = λ . f v=\lambda.f v = λ . f

v = λ f v=\frac{\lambda}{f} v = f λ

Công thức liên hệ vận tốc v v v , bước sóng λ \lambda λ , chu kì T T T

v = λ T v=\frac{\lambda}{T} v = T λ

v = λ . T v=\lambda.T v = λ . T

Điều kiện có cực đại giao thoa

d 2 − d 1 = k λ d_2-d_1=k\lambda d 2 − d 1 = k λ

d 2 − d 1 = ( k + 1 2 ) λ d_2-d_1=\left(k+\frac{1}{2}\right)\lambda d 2 − d 1 = ( k + 2 1 ) λ

Điều kiện có cực tiểu giao thoa

d 2 − d 1 = ( k + 1 2 ) λ d_2-d_1=\left(k+\frac{1}{2}\right)\lambda d 2 − d 1 = ( k + 2 1 ) λ

d 2 − d 1 = k λ d_2-d_1=k\lambda d 2 − d 1 = k λ

Điều kiện có sóng dừng trên sợi dây có hai đầu cố định là:

l = k λ 2 l=k\frac{\lambda}{2} l = k 2 λ

l = ( 2 k + 1 ) λ 4 l=\left(2k+1\right)\ \frac{\lambda}{4} l = ( 2 k + 1 ) 4 λ

Điều kiện có sóng dừng trên sợi dây có một đầu cố định và một đầu tự do là:

l = ( 2 k + 1 ) λ 4 l=\left(2k+1\right)\ \frac{\lambda}{4} l = ( 2 k + 1 ) 4 λ

l = k λ 2 l=k\frac{\lambda}{2} l = k 2 λ

Phương trình sóng tại nguồn O: u O = A cos ⁡ ω t u_O=A\cos\omega t u O = A cos ω t Phương trình sóng tại M cách nguồn O một khoảng x là:

u M = A cos ⁡ 2 π ( t T − x λ ) u_M=A\cos2\pi\left(\frac{t}{T}-\frac{x}{\lambda}\right) u M = A cos 2 π ( T t − λ x )

u M = A cos ⁡ 2 π ( t T + x λ ) u_M=A\cos2\pi\left(\frac{t}{T}+\frac{x}{\lambda}\right) u M = A cos 2 π ( T t + λ x )

Độ lệch pha giữa hai điểm cách nhau một khoảng x trên phương truyền sóng

Δφ = 2 π x λ 2\pi\frac{x}{\lambda} 2 π λ x

Δφ = 2 π λ x 2\pi\ \frac{\lambda}{x} 2 π x λ

Mối liên hệ giữa cường độ dòng điện hiệu dụng và cường độ dòng điện cực đại là:

I=\frac{I_0}{\sqrt{2}} I = 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> I 0

U = U 0 2 U=\frac{U_0}{\sqrt{2}} U = 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> U 0

Mối liên hệ giữa điện áp hiệu dụng và điện áp cực đại là:

U = U 0 2 U=\frac{U_0}{\sqrt{2}} U = 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> U 0

I=\frac{I_0}{\sqrt{2}} I = 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> I 0

Dung kháng của đoạn mạch là:

Z C = 1 ω C Z_C=\frac{1}{\omega C} Z C = ω C 1

Z L = ω L Z_L=\omega L Z L = ω L

Cảm kháng của đoạn mạch là:

Z L = ω L Z_L=\omega L Z L = ω L

Z C = 1 ω C Z_C=\frac{1}{\omega C} Z C = ω C 1

Độ lệch pha của u so với i là

tanφ = Z L − Z C R \frac{Z_L-Z_C}{R} R Z L − Z C

tanφ = Z C − Z L R \frac{Z_C-Z_L}{R} R Z C − Z L

Công thức máy biến áp

N 1 N 2 = U 1 U 2 = I 2 I 1 \frac{N_1}{N_2}=\frac{U_1}{U_2}=\frac{I_2}{I_1} N 2 N 1 = U 2 U 1 = I 1 I 2

N 1 N 2 = I 1 I 2 = U 2 U 1 \frac{N_1}{N_2}=\frac{I_1}{I_2}=\frac{U_2}{U_1} N 2 N 1 = I 2 I 1 = U 1 U 2

Mạch điện xoay chiều cộng hưởng khi

Z L ≠ Z C Z_L\ne Z_C Z L  = Z C

Công thức Tần số của dòng điện do máy phát điện tạo ra?

f = n . p v ớ i n : l a ˋ s o ^ ˊ v o ˋ n g t r e ^ n g i a ^ y f=n.p\ \ \ với\ n:\ là\ số\ vòng\ trên\ giây f = n . p v ớ i n : l a ˋ s o ^ ˊ v o ˋ n g t r e ^ n g i a ^ y

f = n . p v ớ i n : l a ˋ s o ^ ˊ v o ˋ n g t r e ^ n p h u ˊ t f=n.p\ \ \ \ \ với\ n:là\ số\ vòng\ trên\ phút f = n . p v ớ i n : l a ˋ s o ^ ˊ v o ˋ n g t r e ^ n p h u ˊ t

Công thức học kì 1 - Vật lý 12

Authored by Thái Trương

Physics

12th Grade

Used 510+ times

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MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Phương trình dao động điều hòa

x = Acos(ωt + φ)

x = Atsin(ωt+φ)

x = Atan(ωt + φ)

x = Acotan(ωt + φ)

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Trong dao động điều hoà
x = Acos(ωt + φ)
đại lượng A được gọi là

pha ban đầu

chu kì

tần số góc

biên độ

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Trong dao động điều hoà
x = Acos(ωt + φ)
đại lượng ω được gọi là

pha ban đầu

chu kì

tần số góc

biên độ

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Trong dao động điều hoà
x = Acos(ωt + φ)
đại lượng φ được gọi là

pha ban đầu

chu kì

tần số góc

biên độ

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Trong dao động điều hoà
x = Acos(ωt + φ)
vận tốc biến đổi điều hoà theo ptrình

v = Acos(ωt + φ).

v = Aωcos(ωt + φ).

v = - Asin(ωt + φ).

v = - Aωsin(ωt + φ).

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Trong dao động điều hòa
x = Acos(ωt + φ)
tốc độ cực đại được tính theo công thức

v_max = ωA.

v_max = ω²A.

v_max = - ωA.

v_max = - ω²A.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Trong dao động điều hoà
x = Acos(ωt + φ)
gia tốc biến đổi điều hoà theo ptrình

a = Acos(ωt + φ).

a = Aω²cos(ωt + φ).

a = - Aω²cos(ωt + φ).

a = - Aωcos(ωt + φ).

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Trong một dao động bất kì, mối liên hệ giữa tần số góc $\omega$ , chu kì T, tần số f bằng công thức nào?

$\omega=\frac{2\pi}{T}=2\pi.f$

$\omega=\frac{2\pi}{f}=2\pi.T$

$T=\frac{2\pi}{\omega}=2\pi.f$

$T=\frac{2\pi}{f}=2\pi.\omega$

10.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Trong một dao động bất kì, mối liên hệ giữa chu kì T và tần số f là?

$T=\frac{1}{f}$

$T=1.f$

$T=2\pi.f$

$T=\frac{2\pi}{f}$

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Công thức học kì 1 - Vật lý 12

Phương trình dao động điều hòa

Trong dao động điều hoàx = Acos(ωt + φ)đại lượng A được gọi là

Trong dao động điều hoàx = Acos(ωt + φ)đại lượng ω được gọi là

Trong dao động điều hoàx = Acos(ωt + φ)đại lượng φ được gọi là

Trong dao động điều hoà x = Acos(ωt + φ)vận tốc biến đổi điều hoà theo ptrình

Trong dao động điều hòax = Acos(ωt + φ)tốc độ cực đại được tính theo công thức

Trong dao động điều hoà x = Acos(ωt + φ)gia tốc biến đổi điều hoà theo ptrình

Trong dao động điều hòa x = Acos(ωt + φ)giá trị cực đại của gia tốc là

Trong một dao động bất kì, mối liên hệ giữa tần số góc ω\omegaω , chu kì T, tần số f bằng công thức nào?

Trong một dao động bất kì, mối liên hệ giữa chu kì T và tần số f là?

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Popular Resources on Wayground

Discover more resources for Physics

Trong dao động điều hoà
x = Acos(ωt + φ)
đại lượng A được gọi là

Trong dao động điều hoà
x = Acos(ωt + φ)
đại lượng ω được gọi là

Trong dao động điều hoà
x = Acos(ωt + φ)
đại lượng φ được gọi là

Trong dao động điều hoà
x = Acos(ωt + φ)
vận tốc biến đổi điều hoà theo ptrình

Trong dao động điều hòa
x = Acos(ωt + φ)
tốc độ cực đại được tính theo công thức

Trong dao động điều hoà
x = Acos(ωt + φ)
gia tốc biến đổi điều hoà theo ptrình

Trong dao động điều hòa
x = Acos(ωt + φ)
giá trị cực đại của gia tốc là

Trong một dao động bất kì, mối liên hệ giữa tần số góc $\omega$ , chu kì T, tần số f bằng công thức nào?