THEORY: Track down a textbook (or search legitimate internet sites) and review t
ID: 3278704 • Letter: T
Question
THEORY: Track down a textbook (or search legitimate internet sites) and review the theory behind the oscillatory motion of a "simple pendulum" - this is an idealized model of a massless wire of length L terminated by a point mass which, for small angles, oscillates with a period T = 2 pi Squareroot L/g. a) What is the small angle approximation? For how small of angle (in radians!) is it valid to better than 1%? b) A certain pendulum in this solar system has a length of 72.3 (plusminus 0.2) cm and a period of 2.75 (plusminus 0.02) seconds. What is the value of gravity at this location and its uncertainty? What planet is the pendulum located on? Explain.Explanation / Answer
a.
so, for a simple pendulum
when we draw an FBD, assuming that Tension in the string to be T, mass of bob to be m and the position of the string with resperct to the vertical be theta
then from force Balance
T = mgcos(theta) + mv^2/l [ v is speed of the bob and l is length of the string]
mgsin(theta) = ma [ a is acceleration of the bob]
now, a = g*sin(theta)
so if we assume a = g*theta, this becomes a harmonic osscilator with acceleration proportional to the angle of the string in radians
this assumption is called small angle assumption
so let the angle be theta radians
then (sin(theta) - theta)/sin(theta) < 1/100
1 - theta/sin(theta) < 1/100
99/100 < theta/sin(theta)
0.99sin(theta) < theta
sin(theta)/theta) < 1.0101
so theta < 0.25531 rad
b. for a pendulum
time period is given by
T = 2*pisqroot(l/g)
so from error analysis
dT/T = dl/2l + dg/2g
now, T = 2.75 s
dT = 0.02 s
l = 72.3 cm
dl = 0.2 cm
so g = 0.723/(2.75/2*pi)^2 = 3.774 m/s^2
dg = 2*3.774(0.02/2.75 - 0.2/2*72.3) = 0.044
so g = 3.77 +- 0.044 m/s/s
this planet is mass as acceleratoin due to gravity on mars is 3.77 m/s/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.