Considering that the length of the steel is l = 3.80 m and the mass m = 54.0 kg
v = s/t
v = 3.8/0.0492
v = 77.23 m/s
The speed formula for a wire is given by
v = ( T/ÎĽ )^1/2
where T represents the tension in the string and ÎĽ denotes the mass per unit length.
v = ( Tl/m )^1/2.......(substituting ÎĽ with m/l)
Given that T = 550N
v = [ 550(3.8)/m ]^1/2
If we square both sides, we find that
v^2 = 550(3.8)/m
v^2 = 2090/m
Substituting the previously calculated value of v yields the solution
m = 0.35 kg
Quasi frequency = 4√6Quasi period = π√6/12t ≈ 0.4045Explanation: The given data is: Mass, m = 20gτ = 400 dyn.s/cmk = 3920u(0) = 2u'(0) = 0General differential equation:mu" + τu' + ku = 0Substituting the known values yields:20u" + 400u' + 3920u = 0Now, dividing each side by 20 gives:u" + 20u' + 196u = 0To establish the characteristic equation, replace y" with r², y' with r, and y with 1 in the differential equation:r² + 20r + 196 = 0Now, solving for the roots results in:r = -10 ± 4√6iThe general solution for two complex roots is: y = c₁ eᵃt cosbt + c₂ eᵃt sinbtwith a being the real part of the roots and b representing the imaginary part. Given that a = -10 and b = 4√6,u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6twhere u(0) = 2 and u'(0) = 0(b)Quasi frequency: μ = (c)(d)Quasi period: T = 2π / μ(d)|u(t)| < 0.05 cmu(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t| < 0.05solving for t:τ = t ≈ 0.4045
Force: F = m a
a = (v - u)/t, so F = m(v - u)/t
m = mass in kg, v = final velocity in m/s, u = initial velocity in m/s
t = time, Force in newtons.
Given m = 1.2*10³ kg, u = 10 m/s, v = 20 m/s, t = 5 s
F = 1.2*10³(20 - 10)/5
F = 2.4*10³ N = 2400 N
