The notes for the physics lab on simple harmonic motion explore the oscillatory motion of systems like springs and pendulums. Key concepts include amplitude, period, frequency, and the relationship between restoring force and displacement. The lab involves measuring periods and analyzing factors li...
motion question 7
simple harmonic how can you determine ax without massdk ?
·
·
·
when the acceleration of an object is proportional to its displacement a points in the how would you plot
Ta m to obtain linear graph F KAX KDX
Since F mg
·
a
·
mg = =
·
=
,
opposite direction to get a linear graph you would need to square T
bX finda
M this
=
type of periodic motion
/NIm
·
K
units
·
y = mx + b
eg : Suspension spring bungee jumping ↑
·
, ,
·
periodic motion T =
m .
2π n +
c =
y two parameters we don't know
m
you can replace m/k in the period
formula
·
K
any motion that repeats itself after certain time interval
T
-
·
a
↑ X
·
objects that have periodic motion needd restoring force to bring it back to equilibrium slope T =
M2πpT =
P .
2π
the intercept should be zero because when m = 0 , T = 0
·
simple harmonic motion
* however t h e intercept also includes the mass of the pand the now you can rewrite to do a linear ea
.
occurs when
restoring force obeys hooke's law : F -kX
·
=
mass of the Spring .
Therefore while the intercept would ideally why the intercept is not
K: Spring constant
+
+
2 4
be zero ,
in reality it is non-zero becauses these masses are consistent with zero
=
BX
X : the one-dimensional displacement from the
equilibrium position
not physically negligible .
SI units for K : N/m
·
uncertainties
question 2
·
·
question 8
1
·
newton's second law .
m
·
what if the block was hanging vertically ? T = 2π .
F = ma but also F = -kX -
17
FS
,
ma = -
kX F = ma =
mg -
Es
a
Fs kX g
m AM A
q =
question 3
· =
·
mg ma =
mg -
kX
F = max left
1
am
question 9
am
·
= +
F = O m
F = max ·
if the mass is not accelerating , derive equation for X
right Fmax = KA
F = 0
F = max left
ma
: mg-kx 2
.
m
+ a
F =
g Am
F = maX x =
mg
m
right ST
K =
24 -
X is directly proportional to my inversely proportional to k 2
.
m
question S
·
·
the motion of the block can be described as X =
Acos ·
procedures to find K
.
2 k = 442m
A: amplitude maximum ,
distance from equilibrium the box can go ·
procedure 1 : plot T2 vs
. m graph
T2
·
why is T the period of oscillations ?
cosine is an oscillating function that oscillates btwn-1 and 1 every 2it .
analogously
,
&
·
equation : Th =
Th m
A42m =
42 Am .
m
AT = 2 DT T
the the spring oscillates btwn A and A for every period T
.
motion of . -
therefore
you can calculate the spring constant by dividing it by
·
.
M
AK
cosine values repeat every 2 i t
=
i n the same way that the motion repeats every . T2
T
4TRM LBIT
the slope ·
K
4π2
k =
question 6
·
D
m slope
·
what is the equation for T
1x ad F -
F =
ma
procedure
= = ·
2
F =
by measuring the position
-
at equilibrium mg kX we can plot an mg vs X graph
·
= .
,
of the spring when it is not oscillating a placing a different mass on the spring
wixa each time
d
kx w
=
-
=
&
I
wom =
wa K
=
= the slope is k in units of N/m
m &
m
mg & there is no expected bic we are finding it
2π
2
w =
DT =
·
the intercept represents the weight when the spring is
masses are attached
D
T in equilibrium a no
substitute X
) the expected intercept
↳ is O b l we are considering
T = 2π .
m the spring to be massless a frictionless so when
X =
0 , mg =
0
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