To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.
Our values are given as
The angular velocity of a body can be described as a function of frequency as
PART A) The expression for the maximum angular velocity is given by the amplitude so that
PART B) The maximum acceleration on your part would be given by the expression
Answer:
a) 86 atm
b) 86 atm
c) 645 m/s
Explanation:
See attachment for calculations on how i arrived at the answer
Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v =
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ =
with the new dough
v =
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v =
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v =
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v =
v = 1 /√5 v₀
v = 0.447 v₀