Post on 18-Jan-2016
transcript
Equilibrium position
Equilibrium position
xdisplacement
Equilibrium position
x
F
displacement
Equilibrium position
x
F
Resultant force or Restoring force
displacement
Equilibrium position
x
F
Resultant force or Restoring force
displacement
If the resultant force is directed towards and proportional to the displacement from equilibrium,
then so is the acceleration,and the object executes SHM.
If the resultant force is directed towards and proportional to the displacement from equilibrium,
then so is the acceleration,and the object executes SHM.
Effect on the time period of :1. increasing the mass2. using stiffer springs ?
Mass on a spring
Equilibrium position
Downward displacement x
x
m
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F m
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
m
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM
m
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
m
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
m
If the resultant force is directed towards and proportional to the displacement from equilibrium,
then so is the acceleration,and the object executes SHM.
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
When the mass is displaced a small distance xthe resultant upwards restoring force F:
F = - k x
m
If the resultant force is directed towards and proportional to the displacement from equilibrium,
then so is the acceleration,and the object executes SHM.
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
When the mass is displaced a small distance xthe resultant upwards restoring force F:
F = - k x
ma = - k x
m
If the resultant force is directed towards and proportional to the displacement from equilibrium,
then so is the acceleration,and the object executes SHM.
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
When the mass is displaced a small distance xthe resultant upwards restoring force F:
F = - k x
ma = - k x
a = - k x m
m
If the resultant force is directed towards and proportional to the displacement from equilibrium,
then so is the acceleration,and the object executes SHM.
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
When the mass is displaced a small distance xthe resultant upwards restoring force F:
F = - k x - ve sign shows thatfor a downward displacement
there is an upward restoring force!
ma = - k x
a = - k x m
m
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
When the mass is displaced a small distance xthe resultant upwards restoring force F:
F = - k x - ve sign shows thatfor a downward displacement
there is an upward restoring force!
ma = - k x
Compare this with the SHM equation;
a = - k x m
m
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
When the mass is displaced a small distance xthe resultant upwards restoring force F:
F = - k x - ve sign shows thatfor a downward displacement
there is an upward restoring force!
ma = - k x
Compare this with the SHM equation;
a = - k x m
a = - (2πf)2 x
a = - k x m
m
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
When the mass is displaced a small distance xthe resultant upwards restoring force F:
F = - k x - ve sign shows thatfor a downward displacement
there is an upward restoring force!
ma = - k x
Compare this with the SHM equation;
a = - k x m
a = - (2πf)2 x
-k = - (2πf)2 m
a = - k x m
m
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
When the mass is displaced a small distance xthe resultant upwards restoring force F:
F = - k x - ve sign shows thatfor a downward displacement
there is an upward restoring force!
ma = - k x
Compare this with the SHM equation;
a = - k x m
a = - (2πf)2 x
-k = - (2πf)2 m
k = 4 π 2 f 2 m
a = - k x m
m
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
When the mass is displaced a small distance xthe resultant upwards restoring force F:
F = - k x - ve sign shows thatfor a downward displacement
there is an upward restoring force!
ma = - k x
Compare this with the SHM equation;
a = - k x m
a = - (2πf)2 x
-k = - (2πf)2 m
k = 4 π 2 f 2 m
f = 1 k 2π m a = - k x
m
m
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
When the mass is displaced a small distance xthe resultant upwards restoring force F:
F = - k x
ma = - k x
Compare this with the SHM equation;
a = - k x m
a = - (2πf)2 x
-k = - (2πf)2 m
k = 4 π 2 f 2 m
f = 1 k 2π m a = - k x
m
m
T = 2π m k
or :
Mass on a spring
Equilibrium position
Downward displacement x
x
Restoring force F
F
Laws used:Hooke’s law F = k ΔL
Newton’s 2nd F = ma
SHM a = -(2πf)2 x
When the mass is displaced a small distance xthe resultant upwards restoring force F:
F = - k x
ma = - k x
Compare this with the SHM equation;
a = - k x m
a = - (2πf)2 x
-k = - (2πf)2 m
k = 4 π 2 f 2 m
f = 1 k 2π m a = - k x
m
m
T = 2π m k
or :
Mass on a spring
T = 2π m k
Mass on a spring
T = 2π m k
Put in the form: y = m x + c
Mass on a spring
T = 2π m k
Put in the form: y = m x + c
T 2 = 4 π 2 m + 0 k
Mass on a spring
T = 2π m k
Put in the form: y = m x + c
T 2 = 4 π 2 m + 0 k
T 2
/s 2
m / kg
Mass on a spring
T = 2π m k
Put in the form: y = m x + c
T 2 = 4 π 2 m + 0 k
T 2
/s 2
m / kg
Max spring tension = mg + kx
Mass on a spring
T = 2π m k
Put in the form: y = m x + c
T 2 = 4 π 2 m + 0 k
T 2
/s 2
m / kg
Max spring tension = mg + kx
x = A ( amplitude )
Mass on a spring
T = 2π m k
Put in the form: y = m x + c
T 2 = 4 π 2 m + 0 k
T 2
/s 2
m / kg
Max spring tension = mg + kx
Min spring tension = mg - kxx = A ( amplitude )