What is the period of a wave if the wavelength is 100 m and the speed is 200 m/s?
2 answers:
This is simple question and easy to understand.
so first you use a formula to calculate frequency of the wave.
v _ stands for velocity a.k.a speed
f_ stands for frequency which you have to find.
(pronounced lemda)_ stands for wavelength.
Then you gonna place all your values according to you have in the formula as shown then solve for f.
Finally, you use the time formula
where
T_ is period a.k.a time which equal to the reciprocal of frequency or equal to 1/f.
Simply input your values according to the formula and you your answer as 0.5s
so first you use a formula to calculate frequency of the wave.
v _ stands for velocity a.k.a speed
f_ stands for frequency which you have to find.
(pronounced lemda)_ stands for wavelength.
Then you gonna place all your values according to you have in the formula as shown then solve for f.
Finally, you use the time formula
where
T_ is period a.k.a time which equal to the reciprocal of frequency or equal to 1/f.
Simply input your values according to the formula and you your answer as 0.5s
You might be interested in
The frequency of the wave is
Explanation:
The frequency, the wavelength and the speed of a wave are related by the following equation:
where
c is the speed of the wave
f is the frequency
is the wavelength
For the radio wave in this problem,
Therefore, the frequency is:
Learn more about waves here:
#LearnwithBrainly
Answer:
Explanation:
Given
length of rope
velocity while running
when the person jumps off the bank and hang on the rope then we can treat the person as pendulum with Time period T which is given by
Greatest Possible distance will be covered when person reaches the other extreme end of assumed pendulum (velocity=zero)
therefore he must hang on for 0.5 T time
Answer:
the penny loses contact at the piston's highest point.
f = 2.5 Hz
Explanation:
Concepts and Principles
1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by
∑F = ma (1)
2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:
a_max = -w^2A (2)
3- The angular frequency w of a wave is related to the frequency f by:
w = 2π f (3)
Given Data
- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)= 0.04 m.
- The frequency of oscillation of the piston is steadily increased.
Required Data
<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston. </em>
<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle. </em>
Solution
(a)
The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.
figure 1 is attached
Apply Newton's second law from Equation (1) in the vertical direction to the penny:
∑F_y -mg= ma
Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So
0 = m(g + a)
a = —g
Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.
(b)
The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):
a = —w^2A
where a = —g at the highest point. So
g = w^2A
Solve for w:
w =√g/A
Substitute for w from Equation (3):
2πf = √g/A
Solve for f :
f = 1/2π√g/A
Substitute numerical values:
f = 1/2π√9.8 m/s^2/0.04
f = 2.5 Hz
