Problem Deuteronomy Fisikastudycenter.com-wave Daily High School
Physics Class XII Science example problems with the discussion.
b. corner frequency of the wave
c. constant wave
d. wave propagation speed
e. frequency wave
f. wave period
g. wavelength
h. direction of wave propagation
i. deviation of the wave at t = 1 second and x = 1 m
j. wave velocity equation
k. The maximum speed of the wave
l. acceleration of the wave equation
m. absolute value of the maximum acceleration
n. phase angle at t = 0.1 second at x = 1/3 m
o. phase at t = 0.1 second at x = 1/3 m
Wave phase difference between two points is a known distance
a. determines the wavelength
b. determine the frequency of the wave
c. determine the length of rope
Number 4
Given the graph of a traveling wave as shown below!
or
to sign the agreement as follows:
Number 5
A wire vibrates according to the equation:
Third abdominal distance from the point x = 0 is .....
A. 10 cm
B. 7.5 cm
C. 6.0 cm
D. 5.0 cm
E. 2.5 cm
Problem source: Marthen Kanginan 3A Symptoms Wave
Discussion:
Above pattern is the pattern for the wave equation remains stationary end or ends tied. To find the distance from the tip of the abdomen or ikat knot, first determine the value of the wavelength.
Having met wavelength, just enter the formula to find the stomach to -3. Forget the formula, ..!?! Or fear of the opposite and forth with the free end, ..!? Well guys do not have to use the formula, we use an image just like the one below:
Third abdominal position of the tip remains P 3 A is a quarter wavelength or (5/4) λ (A = wave a hill - a valley), so the value of X is:
X = (5/4) λ = (5/4) x 6 cm = 7.5 cm
Number 6
A transverse wave has a frequency of 0.25 Hz. If the distance between two successive points in the wave that has the same phase is 0.125 m, determine the wave propagation speed, expressed in units of cm / s!
Discussion
Data on the matter:
f = 0.25 Hz
Distance of two successive points and phase:
λ = 0, 125 m
ν = .....
ν = λ f
ν = (0.125) (0.25) = 0.03125 m / s = 3.125 cm / s
Number 7
A transverse wave has a frequency of 0.25 Hz. If the distance between two successive points in the wave that has the opposite phase is 0.125 m, determine the wave propagation speed, expressed in units of cm / s!
Discussion
Data on the matter:
f = 0.25 Hz
Distance of two successive points and the opposite phase:
1/2 λ = 0, 125 m → λ = 2 × 0.125 = 0.25 m
ν = .....
ν = λ f
ν = (0.25) (0.25) = 0.0625 m / s = 6.25 cm / s
Number 8
Given a wave equation:
y = 0.05 cos (10t + 2x) meter
Define:
a) Equation speeds
b) the acceleration equation
Discussion
(Y)
↓ derived
(Ν)
↓ derived
(A)
so that:
y = 0.05 cos (10t + 2x) meter
Derivatives are:
ν = - (10) (0.05) sin (10t + 2x)
ν = - 0.5 sin (10t + 2x) m / s
Derivatives are:
a = - (10) (0.5) cos (10t + 2x)
a = - 5 cos (10t + 2x) m / s 2
Source : fisikastudycenter.com
Number 1
Given a wave equation Y = 0.02 sin (10πt - 2πx) with t in seconds, Y and x in meters.
Define:
a. amplitude of the wave Given a wave equation Y = 0.02 sin (10πt - 2πx) with t in seconds, Y and x in meters.
Define:
b. corner frequency of the wave
c. constant wave
d. wave propagation speed
e. frequency wave
f. wave period
g. wavelength
h. direction of wave propagation
i. deviation of the wave at t = 1 second and x = 1 m
j. wave velocity equation
k. The maximum speed of the wave
l. acceleration of the wave equation
m. absolute value of the maximum acceleration
n. phase angle at t = 0.1 second at x = 1/3 m
o. phase at t = 0.1 second at x = 1/3 m
Discussion:
General form of the wave equation:
Y = A sin (ωt - kx)
the A wave amplitude, ω = 2πf and k = 2π / λ is thus:
a. A = 0.02 m
b. ω = 10π rad / s
c. k = 2π
d. v = ω / k = 10π/2π = 5 m / s
c. k = 2π
d. v = ω / k = 10π/2π = 5 m / s
e. f = ω/2π = 10π/2π = 5 Hz
f. T = 1 / f = 1/5 = 0, 2 second
g. λ = 2π / k = 2π/2π = 1 m
h. the direction of the positive x-axis
i. Y = 0.02 sin (10 π-2π) = 0.02 sin (8π) = 0 m
j. v = ω A cos (ωt-kx) = 10π (0.02) cos (10πt-2πx) m / s
k. v max = ωA = 10π (0.02) m / s
l. a =-ω 2 y = - (10π) 2 (0.02) sin (10πt-2πx) m / s 2
m. a max = |-ω 2 A | = |-10π 2 (0.02) | m / s 2
n. phase angle θ = (10.π.0 ,1-2π. (1/3) = 1/3 π = 60 o
o. phase φ = 60 o / 360 o = 1/6
j. v = ω A cos (ωt-kx) = 10π (0.02) cos (10πt-2πx) m / s
k. v max = ωA = 10π (0.02) m / s
l. a =-ω 2 y = - (10π) 2 (0.02) sin (10πt-2πx) m / s 2
m. a max = |-ω 2 A | = |-10π 2 (0.02) | m / s 2
n. phase angle θ = (10.π.0 ,1-2π. (1/3) = 1/3 π = 60 o
o. phase φ = 60 o / 360 o = 1/6
Number 2
A water surface wave whose frequency is 500 Hz propagate with speed of 350 m / s. determine the distance between two different points of phase angle of 60 °!
(Source: Problem SNCA)
A water surface wave whose frequency is 500 Hz propagate with speed of 350 m / s. determine the distance between two different points of phase angle of 60 °!
(Source: Problem SNCA)
Discussion:
First determine the magnitude of the wavelength where
Wave phase difference between two points is a known distance
Number 3
Rope one end is moved up and down while the other end tied. String wave equation is y = 8 sin (0,1 π) π x cos (100t - 12) with y and x in cm and t in units of second. Define:
Rope one end is moved up and down while the other end tied. String wave equation is y = 8 sin (0,1 π) π x cos (100t - 12) with y and x in cm and t in units of second. Define:
a. wavelength
b. frequency wave
c. length of rope
(Source: Problem Ebtanas)
Discussion:
Stationary wave pattern of the above is
a. determines the wavelength
b. determine the frequency of the wave
c. determine the length of rope
Number 4
Given the graph of a traveling wave as shown below!
If the distance of P to Q to within 5 seconds, determine the equation of the wave at the top! (Typical Problem UN)
Discussion:
General form of the wave equation is
or
or
to sign the agreement as follows:
Amplitude sign (+) if the first move towards the top
Amplitude sign (-) if the first movement toward the bottom
Sign in parentheses (+) if the wave propagates in the direction of the negative X axis / left
Sign in parentheses (-) if the wave propagates in the direction of the positive X axis / to the right
grab the data from about wavelength (λ) = 2 meters, and the period (T) =
5/2 second or frequency (f) = 2/5 Hz, enter data into patterns eg 2
used to pattern obtained
Number 5
A wire vibrates according to the equation:
Third abdominal distance from the point x = 0 is .....
A. 10 cm
B. 7.5 cm
C. 6.0 cm
D. 5.0 cm
E. 2.5 cm
Problem source: Marthen Kanginan 3A Symptoms Wave
Discussion:
Above pattern is the pattern for the wave equation remains stationary end or ends tied. To find the distance from the tip of the abdomen or ikat knot, first determine the value of the wavelength.
Having met wavelength, just enter the formula to find the stomach to -3. Forget the formula, ..!?! Or fear of the opposite and forth with the free end, ..!? Well guys do not have to use the formula, we use an image just like the one below:
X = (5/4) λ = (5/4) x 6 cm = 7.5 cm
Number 6
A transverse wave has a frequency of 0.25 Hz. If the distance between two successive points in the wave that has the same phase is 0.125 m, determine the wave propagation speed, expressed in units of cm / s!
Discussion
Data on the matter:
f = 0.25 Hz
Distance of two successive points and phase:
λ = 0, 125 m
ν = .....
ν = λ f
ν = (0.125) (0.25) = 0.03125 m / s = 3.125 cm / s
Number 7
A transverse wave has a frequency of 0.25 Hz. If the distance between two successive points in the wave that has the opposite phase is 0.125 m, determine the wave propagation speed, expressed in units of cm / s!
Discussion
Data on the matter:
f = 0.25 Hz
Distance of two successive points and the opposite phase:
1/2 λ = 0, 125 m → λ = 2 × 0.125 = 0.25 m
ν = .....
ν = λ f
ν = (0.25) (0.25) = 0.0625 m / s = 6.25 cm / s
Given a wave equation:
y = 0.05 cos (10t + 2x) meter
Define:
a) Equation speeds
b) the acceleration equation
Discussion
(Y)
↓ derived
(Ν)
↓ derived
(A)
so that:
y = 0.05 cos (10t + 2x) meter
Derivatives are:
ν = - (10) (0.05) sin (10t + 2x)
ν = - 0.5 sin (10t + 2x) m / s
Derivatives are:
a = - (10) (0.5) cos (10t + 2x)
a = - 5 cos (10t + 2x) m / s 2
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