Topic: Mechanical Waves and Sound Week 5
Learning Sheet 1
I. OBJECTIVE
At the end of the lesson, the learners CAN……
calculate the propagation speed, power transmitted by waves on a string with given tension,
mass, and length.
II. ESSENTIAL KNOWLEDGE AND SKILLS
1.1. Summary of the Essential concept
Speed of Transverse Waves
Consider a wave on a string with a linear mass density (µ) defined as the mass m per unit of length L, as
shown in this equation
m
µ=
L
This string is under tension F that straightens the string when it is distributed. You might have guesses
that increasing the tension easily restores the string in its equilibrium length (or restores it to its original
position), thus increasing the wave speed. Similarly, increasing the linear mass density makes the motion
more languid, thus decreasing the wave speed. It can be shown in a more advanced physics course that the
speed of the wave on the string is
F
v=
√
µ
In this expression, you see that wave speed depends only on two kinds of quantities. The first quantity is
a force that restores the string to equilibrium. The second quantity provides inertia that prevents the string
from returning instantaneously to equilibrium. Thus result can be applied to many kinds of mechanical
waves. Thus, the expression for wave speed has the following general form:
elastic quantity
v=
√ inertial quantity
Example:
Consider a 3.00 kg sphere that hangs in a 2.00 m string with a mass of 1.00 kg. when the end of the
sphere is jerked, a transverse wave travels through the length of the string. What is the speed of the wave at
point a, b and c as shown in the figure here?
Solution:
First, find the mass per unit length of the string.
mstring
µ=
L
1.00 kg
µ=
2.00 m
µ = 0.50 kg/m
For point a, the tensing F is simply the weight of the hanging sphere. So you can easily find the speed of
the wave as follows:
va = F
µ √
v a = msph ere g
√ µ
2
v a = (3.00 kg)(9.8 m/s )
√ 0.50 kg /m
v a = 7.76 m/s
For point b, the tension is the sum of the weights of the hanging sphere and that mass of the spring that
lies below it. Find the mass of the string below point b by multiplying the mass per unit length µ with the
length of the string l b below point b. Thus,
vb = F
√ µ
v b = msph ere g+(µ l b ) g
√m
µ
kg
√ ( )( ) 2
v b = ( 3.00 kg ) 9.8 s 2 +[ 0.50 m (1.50 m)(9.8 m/s )]
0.50 kg/m
v b = 8.57 m/s
For c, the tension is that total of the weights of the sphere and the string. Thus,
vc = F
√ µ
v c = ( m sp h ere+ mstring ) g
√ µ
v c = ( m sp h ere+ mstring ) g
√ µ
2
v c = (3.00 kg+1.00 kg )( 9.8 m/s )
√ 0.50 kg/m
v c = 8.85 m/s
You see that the speed of the wave increases as it goes up.
Standing waves
When a wave comes together (superpose) with another wave, interference occurs. The resulting wave is
the superposition of the individual waves that interfere.
1.2Testing of Knowledge
The wave speed of a wave on a string depends on the tension and the linear mass density. If the tension is
doubled, what happens to the speed of the waves on the string?
1.2. Knowledge Scanning
1. Three things you learned from this topic
2. Two things that are confusing
3. One thing you like the most
1.4. Integration of Learning
Explain do standing wave occur?
III. SELF-DETERMINING PRACTICE
Instructions: Solve the following problems and show your complete solutions.
1. A string with a linear mass density µ = 0.0060 kg/m is tied to the ceiling. A 30 kg mass is tied to
the free end of the string. The string is plucked, sending a pulse down the string. Estimate the
speed of the pulse as it moves down the string.
2. A one-string musical instrument has a 90.0 cm long string with mass 9.00 g. the speed of the
sound in the room where it is being played is 340 m/s. What must be the tension in the string?
IV. REFLECTION
What string instrument do you like playing or would you like to learn? Why?
Reference:
Arevalo, R.L (2017). General Physics 1. Makati City: Diwa Senior High School Series
Prepared By: Jerry G. Tubongbanua
