Name: Franchesca Claire Vega
Date Performed: January 1, 2022 Score: /20
Activity No. 6
STANDING WAVE MOTION
INTRODUCTION
If a uniform string of infinite length is subjected to a tension and one of its ends is
given a vibrating motion perpendicular to the string, a moving disturbance in the form of a
wave is generated and travels along the string. The motion of any particle of the string is at
right angles to the undisturbed position of the string and the waves are transverse waves. The
speed v of the wave is given by
v = λf , (1)
where f is the frequency and λ is the wavelength of the wave. The speed v, in which the
transverse wave travels along the string, is given by
v=
, (2)
Where F is the stretching force or tension in the string and m is the mass of the string whose
length is L. When a wave is reflected by a fixed point at the end of the string, the appearance
of the resulting motion gives no evidence of two waves traveling in opposite directions. If the
frequency is sufficiently great so that the eye cannot follow the motion, the string appears
subdivided into a number of segments, as in the figures below. The points that remain always
at rest are known as nodes and the points with maximum fluctuations are called antinodes.
The vibration as a whole is called standing waves.
A string vibrates with any frequency but it will vibrate with maximum amplitude if
the frequency, the length of the string, and the speed of the waves in the string are so related
that standing waves will be set up. Since both ends of the string are fixed in position, the ends
must be nodes in the standing waves; we can observe from Figures 1.1-1.4 that the length L
of the string must be an integral number N of half wavelengths.
In general,
L = n( 2λ ), where n = 1, 2, 3, …. (3)
Combining equations (1) and (2), we obtain
f = λ1
(4)
�
2�
(5)
or
f=
Solving for F in terms of λ in Eqn. (4), we get
� 2 2
F=
� λ (6)
�
In this experiment, the waves will be generated by an electrically driven vibrator in a
string under tension. The applied tension is varied to produce standing waves of different
wavelengths in the string that makes it vibrate in one, two or more loops. Using the obtained
magnitudes of the wavelength of the standing waves, the tension in the string and mass per
unit length of the string, the frequency of the oscillations produced by the vibrator will be
determined.
OBJECTIVE
To investigate the properties of standing waves in a string using a mechanical vibrator.
APPARATUS
Mechanical vibrator, 60-Hz frequency generator, iron stand, spring balance, meter
stick, string, extension cord
PROCEDURE:
1. Set up the apparatus as shown in Fig. 1.5.
2. Record the mass density of the string as μ = 2.4 x 10−4 Kg / m .
3. Measure the effective length L of the string which is the distance from the vibrator to the
spring balance. Next determine the mass of the vibrating string m. To do this, multiply L byμ.
m = (L x μ) = 2.148 x 10
L = 89.5 cm or 0.895 m
−4
kg
4. Turn on the vibrator and wave generator.
5. Pull the spring balances and adjust the tension of the string so that it vibrates in one
segment (n=1) with maximum amplitude. Record the tension in the table.
6. Repeat step 5 and find the tensions necessary for the string to vibrate in two, three and four
segments (n = 2, 3, 4), keeping the mass m and length L constant. Record the corresponding
tensions in the table.
DATA AND COMPUTATIONS
1. Calculate the wavelength of each wave for each corresponding tension using Eqn.(3).
(4 points)
𝑆𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝜆
𝐿 = 0.895 𝑚
𝑤ℎ𝑒𝑛 𝑛 = 1,
𝜆=
2𝐿
𝑛
= 2 (0.895𝑚)
𝜆 = 1.79 𝑚
𝑤ℎ𝑒𝑛 𝑛 = 2,
𝜆=
2 (0.895𝑚)
2
𝜆 = 0.895 𝑚
𝑤ℎ𝑒𝑛 𝑛 = 3,
𝜆=
2 (0.895𝑚)
3
𝜆 = 0.597 𝑚
𝑤ℎ𝑒𝑛 𝑛 = 4,
𝜆=
2 (0.895𝑚)
4
𝜆 = 0.4475 𝑚
2. Fill up Table 6.1.
n
F (Tension, Wavelength
Y)
(λ)
X = (λ
)
2
2
2
4
1
2.775 N
1.79 m
3.2041 �
2
0.725 N
0.895 m
0.801025 �
3
0.3 N
0.597 m
0.356409 �
4
0.175 N
0.4475 m
0.20025625 �
Σ�=
2
0.6416410506 �
2
0.1270273753 �
2
2
4.56179025 �
0.580743125
2
N�
4
0.1069227
2
N�
4
0.03504484375
2
N�
��=
Σ�=
2
4
0.04010256566 �
�
3.975 N
8.8913775
2
N�
10.26625681 �
Σ�=
�
XY
�
� �
�
4
11.0750278 �
9.614088169
2
N�
3. Using linear regression, compute the slope. In your computations, use tension as y-variable,
λ^2 as x-variable and N as the number of loops. (Use the back page if you need more space for
the computation.) (3 points)
𝑚=
𝑚=
𝑛 (∑ 𝑥𝑖 𝑦𝑖 ) − (∑ 𝑦𝑖 )(∑ 𝑥𝑖 )
𝑛(∑ 𝑥𝑖 2 ) − (∑ 𝑥𝑖 )2
4 (9614088169) − (3975)(4.56179025)
4(11.0750278) − (4.56179025)2
𝑚 = 0.8651800727
~𝑚 = 0.87
4. Solve for f using slope = mf ^2 / L and compare this with the frequency of vibrator. (3 points)
𝑆𝑙𝑜𝑝𝑒 = 0.87
𝑚𝑎𝑠𝑠 = 2.148𝑥10−4 𝑘𝑔
𝐿 = 0.895 𝑚
𝑆𝑙𝑜𝑝𝑒 =
𝑚𝑓 2
𝐿
𝑠𝑙𝑜𝑝𝑒 (𝐿)
√𝑓 2 = √
𝑚
0.87 (0.895 𝑚)
𝑓= √
2.148 𝑥 10−4 𝑘𝑔
𝑓 = 60.20797289 𝐻𝑧
% 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = |
= |
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑣𝑎𝑙𝑢𝑒 − 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
| 𝑥 100%
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑣𝑎𝑙𝑢𝑒
60 𝐻𝑧 − 60.20797289 𝐻𝑧
| 𝑥 100%
60 𝐻𝑧
= 0.3466214899%
~0.35%