Answer:
Explanation:
ω = √k/m = √(33.9/0.28) = 11 rad/s
(a) maximum speed of the oscillating mass
vmax = ωA = 11(0.05) = 0.55 m/s
(b) speed of the oscillating mass when the spring is compressed 1.5 cm
The portion of total energy that is not spring potential is kinetic
½kA² - ½kx² = ½mv²
v = √(k(A² - x²)/m) = √(33.9(0.05² - 0.015²)/0.28 = 0.52482... ≈ 0.52 m/s
(c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position
Different wording, but same question as part (b) 0.52 m/s
(d) value of x at which the speed of the oscillating mass is equal to one-half the maximum value m
The portion of total energy that is not kinetic is spring potential
½kA² - ½mv² = ½kx²
x = √(kA² - m(vmax/2)²) / k) = √(33.9(0.05²) - 0.28(0.55/2)²) / 33.9)
x = 0.043305...≈ 4.3 cm
Answer:
Explanation:
In equilibrium , weight of mug is equal to restoring force .
mg = kx where m is mass of mug , k is spring constant and x is extension .
k / m = g / x = 9.8 ms⁻² / .025 m
= 392
frequency of oscillation n =
= 4.46 per second.
A i checked hope i was right
Answer:
I don't know I'm sorry I will tell you another answer asks me