The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.
The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)
−gsinθ=lθ¨−gainθ=lθ¨
For small angles, θ≪1,θ≪1, and hence sinθ≈θsinθ≈θ. Hence,
θ¨=−glθθ¨=−glθ
This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.
For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause
Answer:
' In layman's terms, the k variable in Hooke's law (F = -kx) indicates stiffness and strength. The higher the value of k, the more force is needed to stretch an object to a given length. This value is dependent on the object's material alongside its shape and Proportons
Explanation:
The object that is subjected to a steady centripetal force moves in a circular motion. If this force suddenly disappear, the object moves in a direction tangential to the circle with respect to the last direction of the centripetal force.
Answer:
(a) right
(b) λ = 4.65m
(c) v = 1519m/s
(d) f = 327Hz
Explanation:
The characteristic sinusoidal pressure fluctuation equation for a wave traveling to the right is given as
P(x,t) = BkASin(kx-ωt)
Where
B = Bulk modulus
k = wave number
A = displacement amplitude
x = position on the x-axis
ω = angular frequency
t = time t
(a) the wave is travelling to the right
(b) from the given equation
ΔP(x,t) = (24.0 / r) sin(1.35r - 2050t)
1.35 compares with k
So k = 1.35rad/m
And k = 2π/ λ
λ = 2π/k = 2π/1.35 = 4.65m
(c) ω =kv
From the given equation ω = 2050 rad/s
v = ω/k = 2050/1.35 = 1519m/s
(d) f = v/λ =1519/4.65 = 327Hz
The average distance between the Earth and the Sun is 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km<span> (584 million mi).</span>