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WAVES
By Michael Lasaleta 39607122
PULSE SPEED ON A
HANGING STRING
A young physics student at UBC had finished his homework early in the
morning and realized he had the rest of the afternoon free. Unfortunately,
he was going through some self-esteem issues and wanted to spend the
afternoon with someone who found him attractive. He ventured campus
in hopes of finding a life long companion that would reassure him of his
good looks. He came across a vendor who was selling magical lamps. The
young physics student thought it would be a good investment because he
could just wish for someone to find him attractive.
PROBLEM
 The vendor sold the student the lamp but the genie that appeared
refused to grant him his wish without proof that the student understood
the concept of pulse speed on a hanging string. The genie hung the 5kg
lamp to the ceiling with a string. The genie then told the student that the
string had a linear mass density of 30g/m and a length of 10m. The genie
then generated a pulse at the lower end of the string and asked, “What is
the speed of the pulse at the lower end, middle, and at the top of the
string? Also, where is the pulse speed greatest and why?”
KNOWN QUANTITIES
Linear mass density of the string, u
30.0g/m = 0.0300kg/m
Mass of the lamp, M
5kg
Length of the String L,
10m
SOLUTION
Tension in a string at any point is a result of the weight of the hanging
mass as well as the weight of string below the point. If we allow x to be
the height at a given point from the bottom of the string, the total weight
under point x is equal to Mg + (uxg), which is equivalent to tension at that
height. The following equation is wave speed, v(x), at a height x:
v(x) =
Mg+uxg
u
EVALUATING
At x=0, x= 5 and x=10
x(0) =
Mg
u
=
5.00kgX9.81m / s
2
0.0300kg / m
= 40.4m / s
x(5.0m) =
Mg+5.0ug
u
=
(5.00kg+5.0mX0.0300kg / m)X9.81m / s
2
0.0300kg / m
= 41.0m / s
x(10.0m) =
Mg+10.0ug
u
=
(5.00kg+10mX0.0300kg / m)X9.81m / s
2
0.0300kg / m
= 41.6m / s
RESULT
The boy explained to the genie the concept behind his results.
The pulse speeds up as a result of the tension in the string increasing
with height, therefore the pulse speed is greatest at the top of the
string. This concept applies to any string of nonzero linear mass
density that is hanging vertically. When a pulse is generated at the
bottom, the pulse moves upwards and the speed increases.
Physics 101 lo3

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Physics 101 lo3

  • 2. PULSE SPEED ON A HANGING STRING A young physics student at UBC had finished his homework early in the morning and realized he had the rest of the afternoon free. Unfortunately, he was going through some self-esteem issues and wanted to spend the afternoon with someone who found him attractive. He ventured campus in hopes of finding a life long companion that would reassure him of his good looks. He came across a vendor who was selling magical lamps. The young physics student thought it would be a good investment because he could just wish for someone to find him attractive.
  • 3. PROBLEM  The vendor sold the student the lamp but the genie that appeared refused to grant him his wish without proof that the student understood the concept of pulse speed on a hanging string. The genie hung the 5kg lamp to the ceiling with a string. The genie then told the student that the string had a linear mass density of 30g/m and a length of 10m. The genie then generated a pulse at the lower end of the string and asked, “What is the speed of the pulse at the lower end, middle, and at the top of the string? Also, where is the pulse speed greatest and why?”
  • 4. KNOWN QUANTITIES Linear mass density of the string, u 30.0g/m = 0.0300kg/m Mass of the lamp, M 5kg Length of the String L, 10m
  • 5. SOLUTION Tension in a string at any point is a result of the weight of the hanging mass as well as the weight of string below the point. If we allow x to be the height at a given point from the bottom of the string, the total weight under point x is equal to Mg + (uxg), which is equivalent to tension at that height. The following equation is wave speed, v(x), at a height x: v(x) = Mg+uxg u
  • 6. EVALUATING At x=0, x= 5 and x=10 x(0) = Mg u = 5.00kgX9.81m / s 2 0.0300kg / m = 40.4m / s x(5.0m) = Mg+5.0ug u = (5.00kg+5.0mX0.0300kg / m)X9.81m / s 2 0.0300kg / m = 41.0m / s x(10.0m) = Mg+10.0ug u = (5.00kg+10mX0.0300kg / m)X9.81m / s 2 0.0300kg / m = 41.6m / s
  • 7. RESULT The boy explained to the genie the concept behind his results. The pulse speeds up as a result of the tension in the string increasing with height, therefore the pulse speed is greatest at the top of the string. This concept applies to any string of nonzero linear mass density that is hanging vertically. When a pulse is generated at the bottom, the pulse moves upwards and the speed increases.