Download Wave on a string To measure the acceleration due to gravity on a

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Density of states wikipedia , lookup

Vibration wikipedia , lookup

Inertia wikipedia , lookup

Center of mass wikipedia , lookup

Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup

Length contraction wikipedia , lookup

Classical central-force problem wikipedia , lookup

Modified Newtonian dynamics wikipedia , lookup

Wave packet wikipedia , lookup

Equations of motion wikipedia , lookup

Jerk (physics) wikipedia , lookup

Work (physics) wikipedia , lookup

Seismometer wikipedia , lookup

Kinematics wikipedia , lookup

Mass versus weight wikipedia , lookup

G-force wikipedia , lookup

Centripetal force wikipedia , lookup

Gravity wikipedia , lookup

Transcript
Wave on a string
To measure the acceleration due to gravity on a distant
planet, an astronaut hangs a 0.070 kg ball from the end of a
wire. The wire has a length of 1.5 m and a linear density of
3.1 10-4 kg/m. Using electronic equipment, the astronaut
measures the time for a transverse pulse to travel the length
of the wire and obtains a value of 0.085 s. The mass of the
wire is negligible compared to the mass of the ball. Determine
the acceleration due to gravity.
Solution:
We have the formula connecting the velocity of the transverse wave on the wire, v,
force on the string, FT and the linear density of the wire,  ,
v
FT

-----------------------------------------------(1)
Where the velocity of the transverse wave
length...of ..the..string
L
v
 ----------------------(2)
time..taken..to..travell..the..length..of ..the..string T
The tension in the string is the force of gravity pulling down on the weight,
FT  mg -------------------------------------(3)
where ‘m’ is the mass of the ball and the ‘g’ is the acceleration due to gravitation the
distant planet.
Substituting equation (2) and (3) in equation (1),
L

T
mg

2
mg
L
  

T 
L 
 
g
T  m
2
Given data are,
m=0.070 kg,
 =3.1 10-4 kg/m,
L=1.5 m,
T=0.085 s.
Substituting we get the acceleration due to gravitation the distant planet ,
L 
g  
T  m
2
4
 1.5  3.1  10


 0.085  0.070
2
 1.37914m/s 2