Answer: a) v = ω /k, b) v = - ωAcos( kx −ωt)
Explanation:
y(x,t)=Asin(kx−ωt) defines the wave equation.
a)
We are asked to find wave speed (v)
Recall that v = fλ
From the wave equation above,
k = 2π/ λ where k is the wave number and λ is the wavelength, λ = 2π /k
ω = 2πf where f is the frequency and ω is the angular frequency.
f = ω/ 2π.
By substituting for λ and ω into the wave speed formulae, we have that
v =( ω/ 2π) × (2π /k)
v = ω/k
b)
y(x,t)=Asin(kx−ωt)
The first derivative of y with respect to x give the velocity (vy)
By using chain rule, we have that
v = dy/dt = A cos( kx −ωt) × (−ω)
v = - ωAcos( kx −ωt)
In a non accelerated frame, also called inertial frame, when no additional forces are acting upon the body.
For instance, a scale in free fall will not record the weight, but zero, like the astronauts. But also, if someone is pulling the object or pushing it, the scale will record some other value.
Answer:
1 eV = 1.60 * 10^-19 J work done in accelerating electron thru 1 V
KE (total energy) = 1350 ^ 1 eV (note proton goes from + to -)
KE = 1.60 * 10^-19 * 1350 = 2.16 * 10^-16 Joules
1/2 m v^2 = KE = 2.16 * 10^-16 J
v^2 = 4.32 * 10E-16 / 1.67 * 10-27 = 2.59 * 10^11
v = 5.09 * 10^5 m/s
Answer:
8.80 Hz
Explanation:
The frequency of a loaded spring is given by

where k and m are the spring constant and the mass of the load respectively. The values of these do not change because they are internal properties of the components of the system.
Hence, the frequency of the vertical spring mass does not change and is 8.80 Hz.
On the other hand, the frequency of the simple pendulum is affected because it is given by

where g and l are acceleration due to gravity and length of the pendulum, respectively. It is thus seen that it depends on g, which changes with location. In fact, the new frequency is given by

Answer:
f1 = 12.90 Hz
Explanation:
To calculate the first harmonic frequency you use the following formula for n = 1:

( 1 )
It is necessary that the unist are in meters, then you have:
L: length of the string = 60cm = 0.6m
M: mass of the string = 0.05kg
T: tension on the string = 20 N
you replace the values of L, M and T in the expression (1) for getting f1:

Hence, the first harmonic has a frequency of 12.90 Hz