Energy of Simple Harmonic Oscillators AP Physics 1 FRQ – Exam Style Questions etc.
Energy of Simple Harmonic Oscillators AP Physics 1 FRQ
Unit 7: Oscillations
Weightage : 10-15%
Exam Style Practice Questions,Energy of Simple Harmonic Oscillators AP Physics 1 FRQ
Question
A 2-kilogram block is dropped from a height of 0.45 meter above an uncompressed spring, as shown above. The spring has an elastic constant of 200 newtons per meter and negligible mass. The block strikes the end of the spring and sticks to it.
a. Determine the speed of the block at the instant it hits the end of the spring
b. Determine the force in the spring when the block reaches the equilibrium position
c. Determine the distance that the spring is compressed at the equilibrium position
d. Determine the speed of the block at the equilibrium position
e. Determine the resulting amplitude of the oscillation that ensues
f. Is the speed of the block a maximum at the equilibrium position, explain.
g. Determine the period of the simple harmonic motion that ensues
Answer/Explanation
Ans:
a) Apply energy conservation from top to end of spring using h=0 as end of spring.
U = K mgh = ½ m v2 (9.8)(0.45) = ½ v2 v = 3 m/s
b) At equilibrium the forces are balanced Fnet = 0 Fsp = mg =(2)(9.8) = 19.6 N
c) Using the force from part b, Fsp = k ∆x 19.6 = 200 ∆x ∆x = 0.098 m
d) Apply energy conservation using the equilibrium position as h = 0. (Note that the height at the top position is now
increased by the amount of ∆x found in part c hnew = h+∆x = 0.45 + 0.098 = 0.548 m
Utop = Usp + K (at equil)
mghnew = ½ k ∆x2 + ½ mv2 (2)(9.8)(0.548) = ½ (200)(0.098)2 + ½ (2)(v2) v = 3.13 m/s
e) Use the turn horizontal trick. Set equilibrium position as zero spring energy then solve it as a horizontal problem
where Kequil = Usp(at max amp.) ½ mv2 = ½ k∆x2 ½ (2)(3.13)2 = ½ (200)(A2) A = 0.313 m
f) This is the maximum speed because this was the point when the spring force and weight were equal to each other and the acceleration was zero. Past this point, the spring force will increase above the value of gravity causing an upwards acceleration which will slow the box down until it reaches its maximum compression and stops momentarily.
g) \(T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{2}{200}} = 0.63s\)
Question
An ideal massless spring is hung from the ceiling and a pan suspension of total mass M is suspended from the end of the spring. A piece of clay, also of mass M, is then dropped from a height H onto the pan and sticks to it. Express all
algebraic answers in terms of the given quantities and fundamental constants.
(a) Determine the speed of the clay at the instant it hits the pan.
(b) Determine the speed of the clay and pan just after the clay strikes it.
(c) After the collision, the apparatus comes to rest at a distance H/2 below the current position. Determine the spring constant of the attached spring.
(d) Determine the resulting period of oscillation.
Answer/Explanation
Ans:
a) Apply energy conservation Utop = Kbot mgh = ½ mv2 v = \(\sqrt{2gH}\)
b) Apply momentum conservation perfect inelastic pbefore = pafter
Mvai= (M+M) vf \(M(\sqrt{2gH}) = 2Mv_{f}\) \(v_{f} = \frac{1}{2}\sqrt{2gH}\)
c) Again we cannot use the turn horizontal trick because we do not know information at the equilibrium position. While the tray was initially at its equilibrium position, its collision with the clay changed where this location would be. Even though the initial current rest position immediately after the collision has an unknown initial stretch to begin with due to the weight of the tray and contains spring energy, we can set this as the zero spring energy position and use the additional stretch distance H/2 given to equate the conversion of kinetic and gravitational energy after the collision into the additional spring energy gained at the end of stretch.
Apply energy conservation K + U = Usp (gained) ½ mv2 + mgh = ½ k ∆x2
Plug in mass (2m), h = H/2 and ∆x = H/2 → ½ (2m)v2 + (2m)g(H/2) = ½ k(H/2)2
plug in vf from part b m(2gH/4) + mgH = kH2/8 ….
Both sides * (1/H) → mg/2 + mg = kH/8 → 3/2 mg = kH/8 k = 12mg / H
d) Based on \(T = 2\pi \sqrt{\frac{2M}{\frac{12 Mg}{H}}} = 2\pi \sqrt{\frac{H}{6g}}\)