Standing waves are created by two harmonic waves of equal amplitude, wavelength and frequency moving in opposite directions. Nodes occur where the amplitude is always zero, while antinodes occur where the amplitude is at its maximum between -2A and 2A. The distance between nodes and antinodes is λ/2, while the distance between a node and the nearest antinode is λ/4. The frequency of the mth harmonic is given by fm = m*f1, where f1 is the fundamental frequency.
My Learning object describes what standing waves are, how to determine where the nodes and antinodes of a standing wave are and also about the fundamental and resonant frequencies. Their is a variety of questions from multiple choice, to true and false and also a problem solving question.
My Learning object describes what standing waves are, how to determine where the nodes and antinodes of a standing wave are and also about the fundamental and resonant frequencies. Their is a variety of questions from multiple choice, to true and false and also a problem solving question.
A presentation covering some of the basics of standing waves and the differences between standing and travelling waves. Practice questions are included at the end.
Please download the presentation in order to see the gifs.
Physics 101 LO6 which explains the components of standing waves, generates its equation, and tests the understanding of students by creating a practice problem with a worked solution in the end.
In this presentation, I explain what a standing wave on a string is, the difference between a standing wave and a travelling wave, and go over some practice problems.
For my LO, I gave some general equations which can be applied for standing wave questions and explained about the pattern of first, second, third harmonic. Then, I gave examples of different standing wave types found in instruments by clarinet and flute. I used a powerpoint to display my learning objective.
CBSE Physics/ Lakshmikanta Satapathy/ Wave Motion Theory/ Reflection of waves/ Traveling and stationary waves/ Nodes and anti-nodes/ Stationary waves in strings/ Laws of transverse vibration of stretched strings
This Learning Object 6 is based on Standing Waves. It gives a detailed description about what a standing wave is and how standing waves are formed. As well, the equations necessary to break down standing waves to get the amplitude, nodes, antinodes, etc are all included. Furthermore, a practice question is included along with a detailed solution to solve the problem with ease.
A presentation covering some of the basics of standing waves and the differences between standing and travelling waves. Practice questions are included at the end.
Please download the presentation in order to see the gifs.
Physics 101 LO6 which explains the components of standing waves, generates its equation, and tests the understanding of students by creating a practice problem with a worked solution in the end.
In this presentation, I explain what a standing wave on a string is, the difference between a standing wave and a travelling wave, and go over some practice problems.
For my LO, I gave some general equations which can be applied for standing wave questions and explained about the pattern of first, second, third harmonic. Then, I gave examples of different standing wave types found in instruments by clarinet and flute. I used a powerpoint to display my learning objective.
CBSE Physics/ Lakshmikanta Satapathy/ Wave Motion Theory/ Reflection of waves/ Traveling and stationary waves/ Nodes and anti-nodes/ Stationary waves in strings/ Laws of transverse vibration of stretched strings
This Learning Object 6 is based on Standing Waves. It gives a detailed description about what a standing wave is and how standing waves are formed. As well, the equations necessary to break down standing waves to get the amplitude, nodes, antinodes, etc are all included. Furthermore, a practice question is included along with a detailed solution to solve the problem with ease.
Learning Object- Standing Waves on Stringskendrick24
This is my learning object about standing waves on a string. I talk about the harmonics, the equation for calculating the frequency for a wave on a string, and gave an example problem.
MAHARASHTRA STATE BOARD
CLASS XI and XII
CHAPTER 6
SUPERPOSITION OF WAVES
CONTENT:
Introduction
Transverse and
longitudinal waves
Displacement relation in a
progressive wave
The speed of a travelling
wave
The principle of
superposition of waves
Reflection of waves
Beats
Doppler effect
This LO gives you a simple easy to understand explanation of what a standing wave is (video included) and how it is different from a travelling wave. Afterwards a few sample questions are given to apply knowledge.
Professional air quality monitoring systems provide immediate, on-site data for analysis, compliance, and decision-making.
Monitor common gases, weather parameters, particulates.
A brief information about the SCOP protein database used in bioinformatics.
The Structural Classification of Proteins (SCOP) database is a comprehensive and authoritative resource for the structural and evolutionary relationships of proteins. It provides a detailed and curated classification of protein structures, grouping them into families, superfamilies, and folds based on their structural and sequence similarities.
(May 29th, 2024) Advancements in Intravital Microscopy- Insights for Preclini...Scintica Instrumentation
Intravital microscopy (IVM) is a powerful tool utilized to study cellular behavior over time and space in vivo. Much of our understanding of cell biology has been accomplished using various in vitro and ex vivo methods; however, these studies do not necessarily reflect the natural dynamics of biological processes. Unlike traditional cell culture or fixed tissue imaging, IVM allows for the ultra-fast high-resolution imaging of cellular processes over time and space and were studied in its natural environment. Real-time visualization of biological processes in the context of an intact organism helps maintain physiological relevance and provide insights into the progression of disease, response to treatments or developmental processes.
In this webinar we give an overview of advanced applications of the IVM system in preclinical research. IVIM technology is a provider of all-in-one intravital microscopy systems and solutions optimized for in vivo imaging of live animal models at sub-micron resolution. The system’s unique features and user-friendly software enables researchers to probe fast dynamic biological processes such as immune cell tracking, cell-cell interaction as well as vascularization and tumor metastasis with exceptional detail. This webinar will also give an overview of IVM being utilized in drug development, offering a view into the intricate interaction between drugs/nanoparticles and tissues in vivo and allows for the evaluation of therapeutic intervention in a variety of tissues and organs. This interdisciplinary collaboration continues to drive the advancements of novel therapeutic strategies.
Slide 1: Title Slide
Extrachromosomal Inheritance
Slide 2: Introduction to Extrachromosomal Inheritance
Definition: Extrachromosomal inheritance refers to the transmission of genetic material that is not found within the nucleus.
Key Components: Involves genes located in mitochondria, chloroplasts, and plasmids.
Slide 3: Mitochondrial Inheritance
Mitochondria: Organelles responsible for energy production.
Mitochondrial DNA (mtDNA): Circular DNA molecule found in mitochondria.
Inheritance Pattern: Maternally inherited, meaning it is passed from mothers to all their offspring.
Diseases: Examples include Leber’s hereditary optic neuropathy (LHON) and mitochondrial myopathy.
Slide 4: Chloroplast Inheritance
Chloroplasts: Organelles responsible for photosynthesis in plants.
Chloroplast DNA (cpDNA): Circular DNA molecule found in chloroplasts.
Inheritance Pattern: Often maternally inherited in most plants, but can vary in some species.
Examples: Variegation in plants, where leaf color patterns are determined by chloroplast DNA.
Slide 5: Plasmid Inheritance
Plasmids: Small, circular DNA molecules found in bacteria and some eukaryotes.
Features: Can carry antibiotic resistance genes and can be transferred between cells through processes like conjugation.
Significance: Important in biotechnology for gene cloning and genetic engineering.
Slide 6: Mechanisms of Extrachromosomal Inheritance
Non-Mendelian Patterns: Do not follow Mendel’s laws of inheritance.
Cytoplasmic Segregation: During cell division, organelles like mitochondria and chloroplasts are randomly distributed to daughter cells.
Heteroplasmy: Presence of more than one type of organellar genome within a cell, leading to variation in expression.
Slide 7: Examples of Extrachromosomal Inheritance
Four O’clock Plant (Mirabilis jalapa): Shows variegated leaves due to different cpDNA in leaf cells.
Petite Mutants in Yeast: Result from mutations in mitochondrial DNA affecting respiration.
Slide 8: Importance of Extrachromosomal Inheritance
Evolution: Provides insight into the evolution of eukaryotic cells.
Medicine: Understanding mitochondrial inheritance helps in diagnosing and treating mitochondrial diseases.
Agriculture: Chloroplast inheritance can be used in plant breeding and genetic modification.
Slide 9: Recent Research and Advances
Gene Editing: Techniques like CRISPR-Cas9 are being used to edit mitochondrial and chloroplast DNA.
Therapies: Development of mitochondrial replacement therapy (MRT) for preventing mitochondrial diseases.
Slide 10: Conclusion
Summary: Extrachromosomal inheritance involves the transmission of genetic material outside the nucleus and plays a crucial role in genetics, medicine, and biotechnology.
Future Directions: Continued research and technological advancements hold promise for new treatments and applications.
Slide 11: Questions and Discussion
Invite Audience: Open the floor for any questions or further discussion on the topic.
Introduction:
RNA interference (RNAi) or Post-Transcriptional Gene Silencing (PTGS) is an important biological process for modulating eukaryotic gene expression.
It is highly conserved process of posttranscriptional gene silencing by which double stranded RNA (dsRNA) causes sequence-specific degradation of mRNA sequences.
dsRNA-induced gene silencing (RNAi) is reported in a wide range of eukaryotes ranging from worms, insects, mammals and plants.
This process mediates resistance to both endogenous parasitic and exogenous pathogenic nucleic acids, and regulates the expression of protein-coding genes.
What are small ncRNAs?
micro RNA (miRNA)
short interfering RNA (siRNA)
Properties of small non-coding RNA:
Involved in silencing mRNA transcripts.
Called “small” because they are usually only about 21-24 nucleotides long.
Synthesized by first cutting up longer precursor sequences (like the 61nt one that Lee discovered).
Silence an mRNA by base pairing with some sequence on the mRNA.
Discovery of siRNA?
The first small RNA:
In 1993 Rosalind Lee (Victor Ambros lab) was studying a non- coding gene in C. elegans, lin-4, that was involved in silencing of another gene, lin-14, at the appropriate time in the
development of the worm C. elegans.
Two small transcripts of lin-4 (22nt and 61nt) were found to be complementary to a sequence in the 3' UTR of lin-14.
Because lin-4 encoded no protein, she deduced that it must be these transcripts that are causing the silencing by RNA-RNA interactions.
Types of RNAi ( non coding RNA)
MiRNA
Length (23-25 nt)
Trans acting
Binds with target MRNA in mismatch
Translation inhibition
Si RNA
Length 21 nt.
Cis acting
Bind with target Mrna in perfect complementary sequence
Piwi-RNA
Length ; 25 to 36 nt.
Expressed in Germ Cells
Regulates trnasposomes activity
MECHANISM OF RNAI:
First the double-stranded RNA teams up with a protein complex named Dicer, which cuts the long RNA into short pieces.
Then another protein complex called RISC (RNA-induced silencing complex) discards one of the two RNA strands.
The RISC-docked, single-stranded RNA then pairs with the homologous mRNA and destroys it.
THE RISC COMPLEX:
RISC is large(>500kD) RNA multi- protein Binding complex which triggers MRNA degradation in response to MRNA
Unwinding of double stranded Si RNA by ATP independent Helicase
Active component of RISC is Ago proteins( ENDONUCLEASE) which cleave target MRNA.
DICER: endonuclease (RNase Family III)
Argonaute: Central Component of the RNA-Induced Silencing Complex (RISC)
One strand of the dsRNA produced by Dicer is retained in the RISC complex in association with Argonaute
ARGONAUTE PROTEIN :
1.PAZ(PIWI/Argonaute/ Zwille)- Recognition of target MRNA
2.PIWI (p-element induced wimpy Testis)- breaks Phosphodiester bond of mRNA.)RNAse H activity.
MiRNA:
The Double-stranded RNAs are naturally produced in eukaryotic cells during development, and they have a key role in regulating gene expression .
Nutraceutical market, scope and growth: Herbal drug technologyLokesh Patil
As consumer awareness of health and wellness rises, the nutraceutical market—which includes goods like functional meals, drinks, and dietary supplements that provide health advantages beyond basic nutrition—is growing significantly. As healthcare expenses rise, the population ages, and people want natural and preventative health solutions more and more, this industry is increasing quickly. Further driving market expansion are product formulation innovations and the use of cutting-edge technology for customized nutrition. With its worldwide reach, the nutraceutical industry is expected to keep growing and provide significant chances for research and investment in a number of categories, including vitamins, minerals, probiotics, and herbal supplements.
THE IMPORTANCE OF MARTIAN ATMOSPHERE SAMPLE RETURN.Sérgio Sacani
The return of a sample of near-surface atmosphere from Mars would facilitate answers to several first-order science questions surrounding the formation and evolution of the planet. One of the important aspects of terrestrial planet formation in general is the role that primary atmospheres played in influencing the chemistry and structure of the planets and their antecedents. Studies of the martian atmosphere can be used to investigate the role of a primary atmosphere in its history. Atmosphere samples would also inform our understanding of the near-surface chemistry of the planet, and ultimately the prospects for life. High-precision isotopic analyses of constituent gases are needed to address these questions, requiring that the analyses are made on returned samples rather than in situ.
Cancer cell metabolism: special Reference to Lactate PathwayAADYARAJPANDEY1
Normal Cell Metabolism:
Cellular respiration describes the series of steps that cells use to break down sugar and other chemicals to get the energy we need to function.
Energy is stored in the bonds of glucose and when glucose is broken down, much of that energy is released.
Cell utilize energy in the form of ATP.
The first step of respiration is called glycolysis. In a series of steps, glycolysis breaks glucose into two smaller molecules - a chemical called pyruvate. A small amount of ATP is formed during this process.
Most healthy cells continue the breakdown in a second process, called the Kreb's cycle. The Kreb's cycle allows cells to “burn” the pyruvates made in glycolysis to get more ATP.
The last step in the breakdown of glucose is called oxidative phosphorylation (Ox-Phos).
It takes place in specialized cell structures called mitochondria. This process produces a large amount of ATP. Importantly, cells need oxygen to complete oxidative phosphorylation.
If a cell completes only glycolysis, only 2 molecules of ATP are made per glucose. However, if the cell completes the entire respiration process (glycolysis - Kreb's - oxidative phosphorylation), about 36 molecules of ATP are created, giving it much more energy to use.
IN CANCER CELL:
Unlike healthy cells that "burn" the entire molecule of sugar to capture a large amount of energy as ATP, cancer cells are wasteful.
Cancer cells only partially break down sugar molecules. They overuse the first step of respiration, glycolysis. They frequently do not complete the second step, oxidative phosphorylation.
This results in only 2 molecules of ATP per each glucose molecule instead of the 36 or so ATPs healthy cells gain. As a result, cancer cells need to use a lot more sugar molecules to get enough energy to survive.
Unlike healthy cells that "burn" the entire molecule of sugar to capture a large amount of energy as ATP, cancer cells are wasteful.
Cancer cells only partially break down sugar molecules. They overuse the first step of respiration, glycolysis. They frequently do not complete the second step, oxidative phosphorylation.
This results in only 2 molecules of ATP per each glucose molecule instead of the 36 or so ATPs healthy cells gain. As a result, cancer cells need to use a lot more sugar molecules to get enough energy to survive.
introduction to WARBERG PHENOMENA:
WARBURG EFFECT Usually, cancer cells are highly glycolytic (glucose addiction) and take up more glucose than do normal cells from outside.
Otto Heinrich Warburg (; 8 October 1883 – 1 August 1970) In 1931 was awarded the Nobel Prize in Physiology for his "discovery of the nature and mode of action of the respiratory enzyme.
WARNBURG EFFECT : cancer cells under aerobic (well-oxygenated) conditions to metabolize glucose to lactate (aerobic glycolysis) is known as the Warburg effect. Warburg made the observation that tumor slices consume glucose and secrete lactate at a higher rate than normal tissues.
Observation of Io’s Resurfacing via Plume Deposition Using Ground-based Adapt...Sérgio Sacani
Since volcanic activity was first discovered on Io from Voyager images in 1979, changes
on Io’s surface have been monitored from both spacecraft and ground-based telescopes.
Here, we present the highest spatial resolution images of Io ever obtained from a groundbased telescope. These images, acquired by the SHARK-VIS instrument on the Large
Binocular Telescope, show evidence of a major resurfacing event on Io’s trailing hemisphere. When compared to the most recent spacecraft images, the SHARK-VIS images
show that a plume deposit from a powerful eruption at Pillan Patera has covered part
of the long-lived Pele plume deposit. Although this type of resurfacing event may be common on Io, few have been detected due to the rarity of spacecraft visits and the previously low spatial resolution available from Earth-based telescopes. The SHARK-VIS instrument ushers in a new era of high resolution imaging of Io’s surface using adaptive
optics at visible wavelengths.
2. What is a Standing Wave?
A wave that does not appear to travel.
Created by two harmonic waves of equal amplitude, wavelength and
frequency, moving in opposite directions.
3. Nodes and Antinodes
Nodes occur where the amplitude is always zero.
Node Node NodeNode
4. Nodes and Antinodes
Antinodes occur where the amplitude moves between -2A and 2A, the
maximum amplitude. (A – amplitude of constituent harmonic waves)
Antinode
Antinode
5. λ
4
λ
2
3λ
4
λ 5λ
4
λ
2
λ
2
Node to Node
Antinode to Antinode
Nodes and Antinodes
Consecutive nodes and consecutive antinodes are
𝜆
2
(half a wavelength) apart
6. The distance between a node and the nearest antinode is
𝜆
4
.
Nodes and Antinodes
λ
4
λ
2
3λ
4
λ 5λ
4
λ
4
Antinode
to Node
λ
4
Node to
Antinode
7. Frequency of Standing Waves on Strings
𝒇 𝒎 =
𝒗
𝝀 𝒎
=
𝒎
𝟐𝑳
𝒗 =
𝒎
𝟐𝑳
𝑻
𝝁
=
𝒎
𝟐𝑳
𝑻
𝑴
𝑳
m = mode number (1, 2, 3, …) (see next slide)
v = wave speed
𝜆 = wavelength
L = length of string
T = tension of the string
M = mass of string
Recall that 𝜇 is the linear mass density of the string
𝑀
𝐿
.
8. Normal Modes
Represented as n in the diagram
The mode number corresponds to the
number of antinodes on the vibrating string
A mode of 2 is double the frequency of the
fundamental frequency (n = 1)
The relationship between the first harmonic
and other harmonics is represented by this
equation:
𝒇 𝒎 = 𝒎𝒇 𝟏
m is a positive integer > 0
9. Harmonics/Resonant Frequencies
The allowed frequencies represented by the equation:
𝒇 𝒎 = 𝒎𝒇 𝟏
In lab 5, you found the fundamental frequency, the lowest frequency that
results in a single antinode at its maximum frequency.
Knowing the fundamental frequency, we can find the harmonics/resonant
frequencies.
11. #1 - The frequency of the fourth
harmonic is…
A) Same as the frequency of the 2nd harmonic.
B)
4
3
times greater than the frequency of the 3rd harmonic.
C) Triple the frequency of the fundamental frequency.
D) Double the frequency of the 3rd harmonic.
12. The frequency of the fourth harmonic
is…
A) Same as the frequency of the 2nd harmonic.
B)
𝟒
𝟑
times greater than the frequency of the 3rd harmonic.
C) Triple the frequency of the fundamental frequency.
D) Double the frequency of the 3rd harmonic.
𝒂⦁𝒇 𝟑 = 𝒇 𝟒
𝒂⦁𝟑𝒇 𝟏 = 𝟒𝒇 𝟏
𝒂 =
𝟒𝒇 𝟏
𝟑𝒇 𝟏
=
𝟒
𝟑
13. #2 - A guitar string that is plucked produces a
standing wave. Its angular velocity is 5 rad/m.
The amplitude is 8mm. The length the string
vibrating is 40cm. The mass of the string is 0.2g.
The tension of the string is equal to the weight
of a 36kg mass on the moon.
A) Imagine that the standing wave is produced by two harmonic waves of equal
amplitude, wavelength, and frequency travelling opposite directions. What would
be the amplitude of the two constituent waves?
B) Write an equation to represent the amplitude of the standing wave.
C) What is the distance between a node and the nearest antinode?
D) What is the wave speed in the string?
E) What is the fifth harmonic?
14. A guitar string that is plucked produces a standing wave. Its angular velocity is 5
rad/m. The amplitude is 8mm. The length the string vibrating is 40cm. The mass
of the string is 0.2g. The tension of the string is equal to the weight of a 36kg
mass on the moon.
A) Imagine that the standing wave is produced by two harmonic waves of
equal amplitude, wavelength, and frequency travelling opposite directions.
What would be the amplitude of the two constituent waves?
The amplitude of the standing wave is double the amplitude of the
constituent wave.
2A = 8mm
A = 4mm
15. A guitar string that is plucked produces a standing wave. Its angular velocity is
5.0 rad/m. The amplitude is 8.0mm. The length the string vibrating is 40cm. The
mass of the string is 0.2g. The tension of the string is equal to the weight of a
36kg mass on the moon.
B) Write an equation to represent the amplitude of the standing wave.
A(x) = A sin(ωx)
A(x) = (8.0 mm)sin(5.0x)
16. A guitar string that is plucked produces a standing wave. Its angular velocity is 5
rad/m. The amplitude is 8mm. The length the string vibrating is 40cm. The mass
of the string is 0.2g. The tension of the string is equal to the weight of a 36kg
mass on the moon.
C) What is the distance between a node and the nearest antinode?
Let us find the distance between nodes.
Recall from the text (eq 14-44): 𝑥 = 𝑚
𝜆
2
, where m = 0, ±1, ±2, ±3, …
We know that ω = 5rad/m and ω =
2𝜋
𝜆
5 =
2𝜋
𝜆
𝜆 =
2𝜋
5
m = 0 x = 0
m = 1 x =
𝜆
2
=
2𝜋
2(5)
=
𝜋
5
m = 2 x = 𝜆 =
2𝜋
5
(continued on next slide)
17. A guitar string that is plucked produces a standing wave. Its angular velocity is 5
rad/m. The amplitude is 8mm. The length the string vibrating is 40cm. The mass
of the string is 0.2g. The tension of the string is equal to the weight of a 36kg
mass on the moon.
C) What is the distance between a node and the nearest antinode?
The location of the nodes are at: x = 0,
𝜋
5
,
2𝜋
5
, …
The location of the first antinode is at:
𝜆
4
=
2𝜋
(4)5
=
𝜋
10
We can find the distance between any node the nearest antinode by finding
the distance between the first node and antinode.
The first node is at x = 0
The first antinode is at x =
𝜋
10
𝜋
10
- 0 =
𝝅
𝟏𝟎
18. A guitar string that is plucked produces a standing wave. Its angular velocity is 5
rad/m. The amplitude is 8mm. The length the string vibrating is 40cm. The mass of
the string is 0.2g. The tension of the string is equal to the weight of a 36kg mass on
the moon.
D) What is the wave speed in the string?
𝑣 =
𝑇
𝜇
𝜇 =
𝑚 𝑠𝑡𝑟𝑖𝑛𝑔
𝐿
=
0.2 𝑥 10−3 𝑘𝑔
0.4𝑚
=
5 𝑥 10−4 𝑘𝑔
𝑚
𝑇 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 36𝑘𝑔 𝑚𝑎𝑠𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑚𝑜𝑜𝑛
𝑊𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑔 𝑚𝑜𝑜𝑛
𝑔 𝑚𝑜𝑜𝑛 =
1
6
𝑔 𝑒𝑎𝑟𝑡ℎ =
1
6
(9.81
𝑚
𝑠2) = 1.635
𝑚
𝑠2
𝑇 = 𝑚𝑔 𝑚𝑜𝑜𝑛 = 36𝑘𝑔 1.635
𝑚
𝑠2 = 58.86𝑁
𝑣 =
58.86
5 𝑥 10−4 = 𝟑𝟒𝟐
𝒎
𝒔
19. A guitar string that is plucked produces a standing wave. Its angular velocity is 5
rad/m. The amplitude is 8mm. The length the string vibrating is 40cm. The mass of
the string is 0.2g. The tension of the string is equal to the weight of a 36kg mass on
the moon.
E) What is the fifth harmonic?
Find the fundamental frequency first.
𝑣 = 𝜆𝑓
𝑓1 =
𝑣
𝜆
𝑣 = 342.34
𝑚
𝑠
𝜆 = 2𝐿 = 2 0.4𝑚 = 0.8𝑚 𝑎𝑡 𝑓𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑓1 =
342.34
𝑚
𝑠
0.8𝑚
= 427.925Hz
𝑁𝑜𝑤 𝑤𝑒 𝑐𝑎𝑛 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑓𝑖𝑓𝑡ℎ ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝑢𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑓𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦.
𝑓5 = 5𝑓1 = 5 427.925Hz = 𝟐𝟏𝟒𝟎𝐇𝐳
20. Works Cited
Wave on a String PhET simulation (wave diagrams)
https://phet.colorado.edu/en/simulation/wave-on-a-string
Physics 101 Textbook (definitions and equations)
Physics for Scientists and Engineers: An Interactive Approach, 1st
Edition
Robert Hawkes, Javed Iqbal, Firas Mansour, Marina Milner-Bolotin,
and Peter Williams
