Frequency and Period of SHM AP Physics 1 FRQ – Exam Style Questions etc.
Frequency and Period of SHM AP Physics 1 FRQ
Unit 7: Oscillations
Weightage : 10-15%
Exam Style Practice Questions,Frequency and Period of SHM AP Physics 1 FRQ
Question: (7 points, suggested time 13 minutes)
Block P of mass m is on a horizontal, frictionless surface and is attached to a spring with spring constant k. The block is oscillating with period TP and amplitude AP about the spring’s equilibrium position x0. A second block Q of mass 2m is then dropped from rest and lands on block P at the instant it passes through the equilibrium position, as shown above. Block Q immediately sticks to the top of block P, and the two-block system oscillates with period TPQ and amplitude APQ.
(a) Determine the numerical value of the ratio TPQ /T P .
(b) The figure is reproduced above. How does the amplitude of oscillation APQ of the two-block system compare with the original amplitude AP of block P alone?
____ APQ < AP ____APQ = AP ____APQ > AP
In a clear, coherent paragraph-length response that may also contain diagrams and/or equations, explain your reasoning.
Answer/Explanation
Ans:
(a)
\(T_{p} = 2\pi \sqrt{\frac{m}{k}}\)
\(T_{PQ} = 2\pi \sqrt{\frac{3m}{k}}\) \(\frac{T_{PQ}}{T_{P}} = \frac{2\pi \sqrt{\frac{3m}{k}}}{2\pi \sqrt{\frac{m}{k}}} = \frac{\sqrt{\frac{3m}{k}}}{\sqrt{\frac{m}{k}}}\)
\(\frac{\sqrt{\frac{3m}{k}}\sqrt{\frac{m}{k}}}{\frac{m}{k}} = \frac{\frac{\sqrt{3m}}{k}}{\frac{m}{k}} = \sqrt{3}\)
(b)
_×_ APQ < AP
Momentum is conserved during the collision, mv0 = 3m, vf ⇒ v0 = 3vf vf = \(\frac{v_{0}}{3}\)
Final velocity is \(\frac{1}{3}\) of the initial velocity.
KEPQ = \(\frac{1}{2}3m \left ( \frac{v_{0}}{3} \right )^{2} = \frac{1}{6} m{v_{0}}^{2} \) vs, \(\frac{1}{2}m{v_{0}}^{2} = KE_{Q}\)
KE at the equilibrium of blocks Q and P is less than with only Block P.
\(V_{s} = \frac{1}{2}kx^{2}\) \(\frac{1}{6}m{v_{0}}^{2} = \frac{1}{2}kx^{2}\) vs. \(\frac{1}{2}m{v_{0}}^{2} = \frac{1}{2}k{x_{1}}^{2}\)
\(X_{PQ} = v_{0} \sqrt{\frac{m}{3k}}\) \(X_{P} = v_{0} \sqrt{\frac{m}{k}}\)
If all KE is converted to spring potential, block P will compress the spring a greater distance. (XPQ < XP)
Therefore APQ < AP
Question
An ideal spring of unstretched length 0.20 m is placed horizontally on a frictionless table as shown above. One end of the spring is fixed and the other end is attached to a block of mass M = 8.0 kg. The 8.0 kg block is also attached to a massless string that passes over a small frictionless pulley. A block of mass m = 4.0 kg hangs from the other end of the string. When this spring-and-blocks system is in equilibrium, the length of the spring is 0.25 m and the 4.0 kg
block is 0.70 m above the floor.
(a) On the figures below, draw free-body diagrams showing and labeling the forces on each block when the system is in equilibrium.
M = 8.0 kg m = 4.0 kg
(b) Calculate the tension in the string.
(c) Calculate the force constant of the spring.
The string is now cut at point P.
(d) Calculate the time taken by the 4.0 kg block to hit the floor.
(e) Calculate the frequency of oscillation of the 8.0 kg block.
(f) Calculate the maximum speed attained by the 8.0 kg block.
Answer/Explanation
Ans:
a) FBD 
b) Simply isolating the 4 kg mass at rest. Fnet = 0 Ft – mg = 0 Ft = 39 N
c) Tension in the string is uniform throughout, now looking at the 8 kg mass,
Fsp = Ft = k∆x 39 = k (0.05) k = 780 N/m
d) 4 kg mass is in free fall. D = vit + ½ g t2 – 0.7 = 0 + ½ (– 9.8)t2 t = 0.38 sec
e) First find the period. \(T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{8}{780}} = 0.63s\)
… then the frequency is given by f = 1/T = 1.6 Hz
f) The 8 kg block will be pulled towards the wall and will reach a maximum speed when it passes the relaxed length of the spring. At this point all of the initial stored potential energy is converted to kinetic energy
Usp = K ½ k ∆x2 = ½ mv2 ½ (780) (0.05)2 = ½ (8) v2 v = 0.49 m/s