Answer:
T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N
Explanation:
a.)
Period: It is already given in the question "oscillator repeats its motion every 0.372 s".
So T=0.372 s
b)
frequency= f = 1/ T
f = 1/ 0.372
f=2.7 Hz
c).
Angular frequency= w= 2πf
w= 2*π*2.7
w=16.9 rad/s
d)
Spring Constant:
As w=![\sqrt{k/m}](https://tex.z-dn.net/?f=%5Csqrt%7Bk%2Fm%7D)
⇒w²= k/m
⇒k= m*w²
⇒k= 0.628 * 16.9² N/m
⇒k=179.2 N/m
e)
The mass will have maximum speed when it passes through the mean position.
At mean position
Maximum elastic potential energy = Maximum kinetic energy
1/2 k A² = 1/2 m v² ( A is amplitude of oscillation)
⇒ v=![\sqrt{k A^2/m}](https://tex.z-dn.net/?f=%5Csqrt%7Bk%20A%5E2%2Fm%7D)
⇒ v=
\
⇒ v= 8.78 m/s
f)
Maximum force will be exerted on the block when it is at maximum distance.
F= k* A ( A is amplitude of oscillation)
F= 179.2 * 0.27 N
F= 48.4 N
I think is reaction
STIMULUS: In physiology, a stimulus is a detectable change in the internal or external environment. When a stimulus is applied to a sensory receptor, it elicits or influences a reflex via stimulus transduction.
Answer:
At some point on say, the receiving screen, light emanating from the left side of the slit will be out of phase (a difference of 1/2 wavelengths) from light coming from the center of the slit.
Thus for every point that is left of the center of the slit, there will be a point on the right side of the slit that is out of phase,
There will be no light on the screen at that particular point and thus there will be a dark fringe there.
That is the basic explanation for the appearance of dark and bright fringes on the receiving screen.
The force that one massive object exerts to attract another object is called gravity or gravitational pull.
<span>The eastern margin is a convergent boundary subduction zone under the South American Plate and the Andes Mountains, forming the Peru–Chile Trench. The southern side is a divergent boundary with the Antarctic Plate, the Chile Rise, where seafloor spreading permits magma to rise.</span>