A 10-kg block is attached to a spring and both reside on a frictionless, horizon
ID: 1487870 • Letter: A
Question
A 10-kg block is attached to a spring and both reside on a frictionless, horizontal plane. The spring constant is 5 N/m. The pulled 5 cm from equilibrium and released. What will be the maximum speed of the block? What is the frequency of oscillation? How much total energy is possessed by the block-spring system"? If after ten complete oscillations the amplitude of the block is reduced to 2 cm. what is the damping constant of the system? Determine whether the function below satisfies the wave equation. y (x, t) = such (kx) (omega t) An electromagnetic wave travels through free space in the +z direction. It has a frequency of 10^14 Hz. It has an electric field amplitude of 33 V/m in the +y-direction. It has its maximum amplitude at z = 0. Find the wavelength. Write an expression for the electric field component of the wave using actual numbers. What is the magnetic field amplitude and direction? Spaceship A leaves earth at a speed of 0.5c with respect to earth and travels in a straight line towards a distant star. 10 years later. Spaceship B leaves earth on the path as Spaceship A traveling at 0.7c with respect to the earth. In the reference frame of the earth, how long does it take Spaceship B to catch up to Spaceship A? In the reference frame of Spaceship B, how long does it take? A hydrogen atom is in the n = 4 excited state. The electron drops to the n = 2 state and then down to the ground state. What wavelength photons are emitted? The energy of the ground state of a 1 nm infinite square well is equal to the n = 3 state of an infinite square well of width L. Find L. An experiment uses Oxygen-19, which decays to Flourine-19 with a half-life of 26.46 s. At the beginning of the experiment, there are 10.000 oxygen atoms. At the end of the experiment there are 100 oxygen atoms. How long did the experiment take? Write the specific equation showing the decay of O-19 to F-19.Explanation / Answer
.1) The equation for a spring-mass system is
. x(t)= A Cos wt w2 = k/m
We can derive the speed
. v(t) = dv/dt = -Aw Sin wt
The energy of the system is
E = ½ k A2
We write the equation explicit System
. x(t) = 0.05 cos wt w2= 5/10= ½ w= 0.707 rad/s
The maximum speed is when cos = - 1
. vmax = A w vmax= 0.05 0.707 v max= 0.035 m/s
. w = 2f f= w /2 f= 0.707/2 f= 0.1125 Hz
E = ½ 5 0.052 E= 0.00625 J
.3) c=lambda f lambda is the symbol for wavelength
. lambda = c/f lambda = 3 108 /1014 = 3 10-6 m
E(x,t) = Eo Cos (kx-wt)
Eo= 33 V/m
. k = 2/ lambda k= 2/3 10-6 = 2.09 106 m-1 lambda is the symbol for wavelength
. w = 2f w= 2 1014 rad/s
E(z,t)= 33 Cos (2.09 106 z - 2 1014 t) oscillating in y axis
. c= Eo/Bo Bo= Eo/c Bo = 33/ 3 108 =11 10-8 T
B(z,t) = 11 10-8 Cos((2.09 106 z - 2 1014 t) oscillating in x axis
.5) 1/ lambda = RH ( 1/ nf2 – 1/no2 ) RH= 1.097 107 m-1 nf = 2 no= 4
1/lambda= 1.097 107 ( 1/22 – 1/42) = 1.097 107 0.1875 = 0.2056
Lambda = 4.86 10-7 m
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