The position of a 0.5 ?? block oscillating back and forth due to
a spring is given by the equation
? = (5??) cos(?/4?). What is the stiffness of the spring?
The position of a 0.5 ?? block oscillating back and forth due to a spring is...
The position of a 0.5 kg object that is oscillating on an ideal spring is given by the equation x = (10cm)cos(10 t), where t is in seconds. At what position x is the kinetic energy one third of the potential energy at that position?
A mass is oscillating back and forth on a spring without friction, as shown. At which position is the magnitude of the acceleration of the mass a maximum? Position 0 is the relaxed (unstretched) position of the mass. Select one: a. 0 b. acceleration is always constant c. M d. E e. cannot determine from the information given.
A classic spring/block system is moving back and forth on an air track (no friction). The mass of the block is 0.5 kg and the spring constant is 200 N/m. The mass is pulled back from equilibrium and then quickly pushed into motion such that it has an initial speed at the release point (60 cm from the equilibrium position) of 5.0 m/s. Determine the frequency of this oscillator.
Calculate the angular frequency of a mass on a spring which is oscillating back and forth with a period of 0.2 seconds per cycle.
Problem 13. A block oscillating on a spring is subject to a damping force that is proportional to the velocity of the block. The damping force results because the block is sloshing back and forth through a fluid. The amount of damping can be changed by using different damping fluids. For one particular choice of fluid the block is pulled out a distance of 2 m and released. The resulting displacement of the block as a function of time is...
13.6 The equation for the position as a function of time for an oscillating spring is given by x 15 cm cos 47at a) What is the frequency? b) If the mass on the spring is 400 g, what is the spring constant of the spring? c) What is the position at t-0.023 82 d) What is the position at rad 1 0.08 s
A simple harmonic oscillator consists of a block attached to a spring, moving back and forth on a frictionless horizontal surface. Suppose the mass of the box is 5.0 kg. The motion is started by holding the box at 0.50 m from its central position, using a force of 40.0 N. Then the box is let go and allowed to perform simple harmonic motion. roosoo - 5m o +5 m (a) (2 points) What is the amplitude of the motion?...
A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 3.00 Hz, with an amplitude of 5.08 x 10-2m. At the point where the block has its maximum speed, it splits into two identical (equal-mass) blocks and only one of these remains attached to the spring. A. What is the amplitude and frequency of the simple harmonic motion of the piece that remains attached to the spring? B....
t/f with explanations 1) A block oscillates back and forth horizontally on a table due to being attached to spring. The block travels the fastest when the spring is most stretched. Answer 2) A block oscillates back and forth horizontally on a frictionless table due to being attached to spring. The frequency of oscillation would be smaller if a more massive block was used. Answer 3) Two longitudinal waves with different amplitudes are observed. The wave with the greatest intensity...
The position of an air-track cart that is oscillating on a spring is given by the equation x = (12.4 cm) cos[(6.35 s-1)t]. At what value of t after t = 0.00 s is the cart first located at x = 8.47 cm?