Homework 11 Solutions 1. [10 points] A grandfather clock depends on the period of a pendulum to keep correct time. Suppose such a clock is calibrated correctly and then the temperature of the room in which it resides increases. a. Does the clock run slow, fast, or correctly? [Hint: A material expands when its temperature increases.] b. How could we adjust the length of the pendulum to correct the time? Answer: Since a grandfather clock can be pictured as a simple pendulum, it undergoes simple harmonic motion. The period of a simple pendulum is given by √ As the temperature in the room increases, the pendulum expands and thus the period increases. This implies that the grandfather clock will run slower than normal. To fix this problem, one could decrease the room temperature or one could add a small screw to the bottom of the pendulum rod in order to adjust the effective length of the pendulum. 2. [10 points] A block of mass is attached to a spring of force constant that lies on a horizontal surface as shown below. The block is pulled to a position to the right of the equilibrium and released from rest. a. Find the work required to stretch the spring. b. Find the speed the block has as it passes through equilibrium. Answer: The work required to stretch the spring a distance of 5.00 cm is given by ( )( ) As the block passes through equilibrium (at x = 0), the speed of the block is given by √ ( ) √ √ ( ) 3. [10 points] A 2.00-kg object on a frictionless track is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.00 m to the right from its equilibrium position, initiating simple harmonic motion. a. What is the force (magnitude and direction) acting on the object 3.50 s after it is released. b. How many times does the object oscillate in 3.50 s? Answer: (a) To determine the force acting on the object 3.50 s after it’s released, we need to determine the displacement at 3.50 s. We can use the position function of simple harmonic motion to determine this ( ) ( ) (√ ( ) [√ ) ( )] Since the displacement is to the right of its equilibrium position, then the force is directed to the left with magnitude ( ) ( ) ( )( ) (b) To determine the number of times, n, that the object oscillates in 3.50 s, we note that n should be related to the period of oscillation √ √ 4. [10 points] The position of a 0.30-kg object attached to a spring is described by ( ) ( ) a. Find the amplitude of the motion and the spring constant b. Find the position of the object and the force exerted on the object at c. Find the object’s speed and acceleration at d. Determine the kinetic energy, potential energy, and total energy at Answer: (a) Based on the position function, the amplitude of the motion is and the angular frequency of the motion is . The spring constant is given by √ (b) At ) ( ( ) , the position of the object is ( ) ( ) ( ) Therefore, the force exerted on the object is given by ( ) ( (c) The object’s speed at ) )( ( ) is given by ( )( The object’s acceleration at ( ) ( ) is given by )( (d) The kinetic energy of the object at ( ) is given by ( The potential energy of the object at ) )( ) is given by ( Therefore, the total energy is given by )( ) 5. [10 points] A 5.00-g bullet moving with an initial speed of 400 m/s is fired into and passes through a 1.00-kg block, as shown below. The block, initially at rest on a frictionless horizontal surface, is connected to a spring with a spring constant of 900 N/m. If the block moves 5.00 cm to the right after impact, a. Find the speed at which the bullet emerges from the block and b. Find the mechanical energy lost in the collision. Answer: (a) When the bullet is fired into the 1.00-kg block, the block and bullet undergo an inelastic collision. Therefore, momentum is conserved and mechanical energy is lost in the process. Momentum conservation gives Since the block moves 5.00 cm right after impact, the kinetic energy gained by the block after the collision is transferred to elastic potential energy of the spring. We can use mechanical energy conservation to determine the speed of the block after the collision √ √ ( ) Therefore, the speed of the bullet after it emerges from the block is ( )( ) ( ( )( ) ) (b) Since the bullet and block undergo an inelastic collision, there will be a loss of mechanical energy. This energy loss is given as ( [ ( ) )( ) ( )( ( ) ] )( ) Bonus: [6 points] A large block P executes simple harmonic motion as it slides across a frictionless surface with a frequency . Block B rests on it, as shown below, and the coefficient of static friction between the two is . What maximum amplitude of oscillation can the system have if block B is not to slip? Answer: If block B does not slip, then the blocks remain stuck together as the spring is compressed. Therefore, the frictional force acting between them is static friction. Moreover, if both blocks remain stuck together, this implies that block B must have the same acceleration as block P. The horizontal equation of motion for block P is given by ∑ ( ) Since the acceleration of both blocks must be the same, then this implies that ( ) Substituting (2) into (1) gives ( ( ) [ )( ( ) )]

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