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An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounces up and down in
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- The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?arrow_forwardWhich of the following statements is not true regarding a massspring system that moves with simple harmonic motion in the absence of friction? (a) The total energy of the system remains constant. (b) The energy of the system is continually transformed between kinetic and potential energy. (c) The total energy of the system is proportional to the square of the amplitude. (d) The potential energy stored in the system is greatest when the mass passes through the equilibrium position. (e) The velocity of the oscillating mass has its maximum value when the mass passes through the equilibrium position.arrow_forwardIf the amplitude of a damped oscillator decreases to 1/e of its initial value after n periods, show that the frequency of the oscillator must be approximately [1 − (8π2n2)−1] times the frequency of the corresponding undamped oscillator.arrow_forward
- If a simple pendulum oscillates with small amplitude and its length is doubled, what happens to the frequency of its motion? (a) It doubles. (b) It becomes 2 times as large. (c) It becomes half as large. (d) It becomes 1/2 times as large. (e) It remains the same.arrow_forwardShow that angular frequency of a physical pendulum phy=mgrCM/I (Eq. 16.33) equals the angular frequency of a simple pendulum smp=g/, (Eq. 16.29) in the case of a particle at the end of a string of length .arrow_forwardWe do not need the analogy in Equation 16.30 to write expressions for the translational displacement of a pendulum bob along the circular arc s(t), translational speed v(t), and translational acceleration a(t). Show that they are given by s(t) = smax cos (smpt + ) v(t) = vmax sin (smpt + ) a(t) = amax cos(smpt + ) respectively, where smax = max with being the length of the pendulum, vmax = smax smp, and amax = smax smp2.arrow_forward
- A block of mass m = 2.00 kg is attached to a spring of force constant k = 500 N/m as shown in Figure P7.15. The block is pulled to a position xi = 5.00 cm to the right of equilibrium and released from rest. Find the speed the block has as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface is k = 0.350. Figure P7.15arrow_forwardThe total energy of a simple harmonic oscillator with amplitude 3.00 cm is 0.500 J. a. What is the kinetic energy of the system when the position of the oscillator is 0.750 cm? b. What is the potential energy of the system at this position? c. What is the position for which the potential energy of the system is equal to its kinetic energy? d. For a simple harmonic oscillator, what, if any, are the positions for which the kinetic energy of the system exceeds the maximum potential energy of the system? Explain your answer. FIGURE P16.73arrow_forwardA simple harmonic oscillator has amplitude A and period T. Find the minimum time required for its position to change from x = A to x = A/2 in terms of the period T.arrow_forward
- A block of unknown mass is attached to a spring with a spring constant of 6.50 N/m and undergoes simple harmonic motion with an amplitude of 10.0 cm. When the block is halfway between its equilibrium position and the end point, its speed is measured to be 30.0 cm/s. Calculate (a) the mass of the block, (b) the period of the motion, and (c) the maximum acceleration of the block.arrow_forwardIf a car has a suspension system with a force constant of 5.00104 N/m , how much energy must the car’s shocks remove to dampen an oscillation starting with a maximum displacement of 0.0750 m?arrow_forwardWhen a block of mass M, connected to the end of a spring of mass ms = 7.40 g and force constant k, is set into simple harmonic motion, the period of its motion is T=2M+(ms/3)k A two-part experiment is conducted with the use of blocks of various masses suspended vertically from the spring as shown in Figure P15.76. (a) Static extensions of 17.0, 29.3, 35.3, 41.3, 47.1, and 49.3 cm are measured for M values of 20.0, 40.0, 50.0, 60.0, 70.0, and 80.0 g, respectively. Construct a graph of Mg versus x and perform a linear least-squares fit to the data. (b) From the slope of your graph, determine a value for k for this spring. (c) The system is now set into simple harmonic motion, and periods are measured with a stopwatch. With M = 80.0 g, the total time interval required for ten oscillations is measured to be 13.41 s. The experiment is repeated with M values of 70.0, 60.0, 50.0, 40.0, and 20.0 g, with corresponding time intervals for ten oscillations of 12.52, 11.67, 10.67, 9.62, and 7.03 s. Make a table of these masses and times. (d) Compute the experimental value for T from each of these measurements. (e) Plot a graph of T2 versus M and (f) determine a value for k from the slope of the linear least-squares fit through the data points. (g) Compare this value of k with that obtained in part (b). (h) Obtain a value for ms from your graph and compare it with the given value of 7.40 g.arrow_forward
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