Question

A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position yi such that the spring is at its rest length. The object is then released from yi and oscillates up and down, with its lowest position being 16 cm below yi. (a) What is the frequency of the oscillation? Hz (b) What is the speed of the object when it is 15.0 cm below the initial position? m/s (c) An object of mass 275 g is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object? g (d) How far below yi is the new equilibrium (rest) position with both objects attached to the spring? m

Answer 1

a) 1.762Hz, b) 0.43 m/s c) 91.67 g d) 0.32 m

Step-by-step explanation:

given: 2A = 16cm, A = 8 cm, kx = mg where k is the force and g is acceleration due to gravity = 9.81 m/s²

kA = mg

k/m = g /A

F ( frequency ) = 1/T ( T is period) = 1/2π√(k/m)

a) F = 1/(2 × 3.142) × √ (9.81/0.08) = 1.762 Hz

b) potential energy of the body = 0.5 ky² - mgy

where y it the position of the body

kinetic energy of the body = 0.5 mv²

since it conserve

potential energy + kinetic of the body = 0

0.5 ky² - mgy + 0.5 mv² = 0

v = √(2gy - (k/m)y²) = √ ((2×9.81×0.15) - ((9.81/0.08) 0.15²)) = √(2.943 -2.76 ) = 0.43 m/s

c) w = √ (k/(m+M)) = 1/4 √ k/m since the frequency is halfed

k/ (m+M) = 1/4(k/m)

cancel k on both sides

4m - m = M

3m = 275g

m = 275 g/ 3 = 91.67 g

d) ky = (m+M)g

y ( new position) = (m+M)g/k

where k = m(2πf)²

y = (m+M)g / (m(2πf)²) = (91.67 + 275g) *9.81 / ( 5656.13) = 3597.03 / 11238.579 = 0.32 m