Rosie MacLennan is a gold medalist in trampoline gymnastics who credits her success to understanding standing waves. Marco wants to study trampoline physics. When Rosie jumps, she forms standing waves through constructive interference of equal, opposite waves. The document provides information about Rosie's trampoline and asks Marco to analyze the standing waves, accounting for both no damping and damping effects.

1. PHYSICS101
LO6 WEEK 9
JENNYLEE (STUDENT NUMBER 34729146)
THE PHYSICS OF THE TRAMPOLINE:
STANDING WAVES
Rosie MacLennan is a Canadian trampoline gymnast who won a gold medal in
Women’s Trampoline at the 2012 Olympic Games. When asked about her successes, she always thanks
the amazing powers of Physics 101 for helping her to understand the nature of the waves that are
produced with every jump. You are Marco, who wishes to one day participate in the sport of trampoline in
the future Olympic Games. You have decided to study the physics of Rosie’s jumps on the trampoline.
Every time Rosie jumps on the trampoline, she does so in a manner that encourages the
constructive interference of waves that are equal in amplitude, frequency, and wavelength but in
opposite directions. Thus, it is only under these circumstances that we can conclude that each jump
she makes forms standing waves.
WHAT WE KNOW: Through careful lab testing, it was determined that the Olympic standard for a
trampoline has a tension of 72N and a mass density of 0.88 kg/m. The length of the trampoline over
which waves occur is 12.0m. Keep this information in mind, as you will need it to answer the
following questions!
Take the left end of the trampoline to be 0m, the starting point of the wave.
Question One
a) In a typical jump that Rosie makes, the wavelength is measured to be 12.0 meters. In order to
write an equation for the standing wave that Rosie produces, what must be the maximum
amplitude, A and wave number, k if the wave produced by the trampoline has an amplitude of
10.0m at 8.0 meters from the right end of the trampoline? Assume that no damping of the
wave occurs.
Before starting this question, it may be helpful to refer to Example 14-2 in the textbook.
b) Using the equation you determined, find how much the trampoline has lifted/depressed when the
position of the wave is at 7.6m.
c) It is important that Rosie does not land on a node, which yields poor heights. If the anti-nodes of
the wave are ideal positions for Rosie to land on, which locations should she aim to land on the
trampoline? List the ideal locations up to (but not including) 12.0m. Assume that the first node
occurs right at 0m, the left end of the trampoline.

2. PHYSICS101
LO6 WEEK 9
JENNYLEE (STUDENT NUMBER 34729146)
Question Two
Damping is caused by energy loss, and naturally, all waves are subject to this phenomenon,
even the amazing waves that Rosie produces. Since you would like to make your knowledge of standing
waves on a trampoline as applicable as possible, you want to investigate the role of damping in standing
waves. Assume that the trampoline has a damping constant, b, which was measured to be 2.16 kg*s-1,
and a spring constant, k of 98.0 N/m.
Review: The wave speed for damping can be found by 𝜔 𝐷 = √
𝑘
𝑚
− (
𝑏
2𝑚
)2 .
Before starting this question, it may be helpful to refer to Example 13-11 in page 369 of the text.
a) Is the k value used to determine the angular frequency with respect to damping the same as the
k in the function for a standing wave?
b) What would be the new wave speed once we account for damping?
c) How could we have solved for the angular frequency (without damping) without knowing the
spring constant? This will allow you to determine the angular frequency in a standing wave if
you are not given spring constant.
d) How does the angular frequency relate to a standing wave?
ANSWER KEY
QUESTION ONE
PART A
TIP 1: When answering such questions, it is helpful to first list all known variables.
Remember! For a standing wave,
𝐴( 𝑥) = 2𝐴𝑠𝑖𝑛( 𝑘𝑥) = 2𝐴𝑠𝑖𝑛(
2𝜋𝑥
λ
). If you don’t remember where this equation
comes from, or are very confused about how this equation is derived from
the original D(x,t) function, refer to equations (14-38) to (14-42).
First, find the wave number, k.
Be careful! The question indicates that the position is at 8.0m from the
right end of the trampoline. Since the origin is indicated to be the left end of

3. PHYSICS101
LO6 WEEK 9
JENNYLEE (STUDENT NUMBER 34729146)
the trampoline, the question is actually specifying 4.0m from the left side of
the trampoline.
𝑘 =
2𝜋
λ
=
2𝜋
12.0
=
𝜋
6
𝑟𝑎𝑑/𝑚
Second, use the given point in the question as well as the calculated wave number to solve for the
amplitude.
10.0 = 2𝐴𝑠𝑖𝑛 (
𝜋
6
(4.0))
5.0 = 𝐴𝑠𝑖𝑛 (
4𝜋
6
)
𝐴 = 5.77
PART B
TIP 2: Write out the equation that Part A is describing first.
𝐴( 𝑥) = 11.5 sin(
𝜋𝑥
6
)
Now, you can simply plug in the given position to determine at what amplitude the position is 7.6m.
𝐴( 𝑥) = 11.5 sin(
7.6𝜋
6
)
𝐴( 𝑥) = −8.58𝑚.
The amplitude of the wave is -8.58m. Thus, the trampoline has depressed by 8.58m.
PART C
Anti-nodes and nodes occur at
λ
4
apart, while two consecutive anti-nodes are
λ
2
apart. We need to solve for
the first location from the origin that is an anti-node. We are given that the first node occurs right at the
origin; thus, we can conclude the first anti-node is at x =
λ
4
. To find the remaining anti-nodes, solve for
multiples of
λ
4
.
λ =12.0m;
λ
4
= 3.0𝑚. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑖𝑑𝑒𝑎𝑙 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛𝑠 𝑓𝑜𝑟 𝑅𝑜𝑠𝑖𝑒 𝑡𝑜 𝑙𝑎𝑛𝑑 𝑜𝑛 𝑎𝑟𝑒 𝑎𝑡 3.0𝑚, 6.0𝑚, 𝑎𝑛𝑑 9.0𝑚.
QUESTION TWO
PART A
No, the k values for a standing wave and the angular frequency are not the same!
IMPORTANT: Do not confuse the k, spring constant in the damping equation for k, wave number in the
previous equation!
PART B
We are given the damping constant and the spring constant, but we still need to determine the mass of
the trampoline that produces the wave. Since we know the mass density and length, we can solve for the
mass of the trampoline.

4. PHYSICS101
LO6 WEEK 9
JENNYLEE (STUDENT NUMBER 34729146)
𝜇 =
𝑚
𝐿
𝑚 = 𝜇 × 𝐿
𝑚 = 0.88
𝑘𝑔
𝑚
× 12.0𝑚
𝑚 = 10.6𝑘𝑔
Now, we have all the values we need.
Plugging in for the given equation, 𝜔 𝐷 = √ 𝑘
𝑚
− (
𝑏
2𝑚
)
2
𝜔 𝐷 = √
98.0𝑁
𝑚
10.56𝑘𝑔
− (
2.16𝑘𝑔
𝑠
2 × 10.56𝑘𝑔
)
2
𝜔 𝐷 = 9.27𝑟𝑎𝑑/𝑠
We can find the angular frequency, taking into account the damping of the wave.
PART C
If we do not know the value of the spring constant, we must find the angular frequency of the wave by
using the following equations:
𝑣 = √
𝑇 𝑠
𝜇
=
9.05𝑚
𝑠
𝑣 = λf
𝑓 =
𝑣
λ
= 0.754Hz
Since we know that the frequency is equal to the reciprocal of the period, and that 𝑇 =
2𝜋
𝜔
, we can solve
for the angular frequency.
𝜔 = 2𝜋𝑓 = 2𝜋(0.754) = 4.74𝑟𝑎𝑑/𝑠
PART D
Angular frequency in a standing wave is present in the wave function:
, where 2pi/T is equivalent to angular frequency.