Answer:
Δt'/ T% = 90.3%
Explanation:
Simple harmonic movement is described by the expression
x = A cos (wt)
we find the time for the two points of motion
x = - 0.3 A
-0.3 A = A cos (w t₁)
w t₁ = cos -1 (-0.3)
remember that angles are in radians
w t₁ = 1.875 rad
x = 0.3 A
0.3 A = A cos w t₂
w t₂ = cos -1 (0.3)
w t₂ = 1,266 rad
Now let's calculate the time of a complete period
x= -A
w t₃ = cos⁻¹ (-1)
w t₃ = π rad
this angle for the forward movement and the same time for the return movement in the oscillation to the same point, which is the definition of period
T = 2 t₃
T = 2π / w s
now we can calculate the fraction of time in the given time interval
Δt / T = (t₁ -t₂) / T
Δt / T = (1,875 - 1,266) / 2pi
Δt / T = 0.0969
This is the fraction for when the mass is from 0 to 0.3, for regions of oscillation of greater amplitude the fraction is
Δt'/ T = 1 - 0.0969
Δt '/ T = 0.903
Δt'/ T% = 90.3%
Answer:
BC and DE
Explanation:
In the given figure, the velocity time graph is shown. We know that the area under v-t curve gives the displacement of the particle.
Area under AB, 
Area under BC, 
Area under CD, 
Area under DE, 
Area under EF, 
So, form above calculations it is clear that, during BC and DE undergo equal displacement. Hence, the correct option is (c) "BC and DE = 4 meters".
Answer:
You were a freeloader of my questions, so I'll be one too.
