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You hang an object onto a vertically hanging spring and measure the stretch length of the spring to be 1 meter. You then pull down on the object and release.

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Presentation on theme: "You hang an object onto a vertically hanging spring and measure the stretch length of the spring to be 1 meter. You then pull down on the object and release."— Presentation transcript:

1 You hang an object onto a vertically hanging spring and measure the stretch length of the spring to be 1 meter. You then pull down on the object and release it so that it oscillates in simple harmonic motion. The period of this oscillation will be a) about half a second, b) about 1 second, c) about 2 seconds, or d) impossible to determine without knowing the mass or spring constant. (end of section 15.2) QUICK QUIZ 15.1

2 (c). This problem illustrates an easy method for determining the properties of a spring-object system. When you hang the object, the spring force, kx, will be equal to the weight, mg, so that kx = mg or x/g = m/k. From Equation 15.13, QUICK QUIZ 15.1 ANSWER

3 You hang an object onto a vertically hanging spring. You slowly let the object down and it stretches the spring a distance d 1. You then remove the object, return the spring to its equilibrium position and reattach the object. This time, instead of slowly letting the object down, you release it all at once and measure the distance the spring stretches, d 2, before the object springs back up again. The ratio d 2 /d 1 will be a) 1, b) 2, c) 4, or d) impossible to determine without knowing the mass and spring constant. (end of section 15.3) QUICK QUIZ 15.1

4 (b). When you slowly let the object down, the object is in static equilibrium and therefore the spring force, kx, is equal to the weight, mg, so that x = d 1 = mg/k. When you quickly release the mass, the mass will oscillate between its release point and its lowest point of travel. The initial gravitational potential energy of the mass is converted to spring potential energy when the mass is at its lowest point. Therefore, mgx = kx 2 /2, so that x = d 2 = 2mg/k. QUICK QUIZ 15.2 ANSWER

5 You release a pendulum bob from an angle which is 5° from the vertical and measure its period of oscillation. You then repeat the experiment but from an angle which is 90° from the vertical. The period of oscillation for the pendulum in the second case will be a) greater than the first case, b) less than the first case, c) the same as the first case, d) impossible to determine. (end of section 15.5) QUICK QUIZ 15.3

6 (a) and (d). Equation for the simple pendulum is valid for small angles when sin ~. Only then is the period independent of the amplitude. For = 5° or radians, sin is and Equation is clearly valid. For = 90° or 1.57 radians, sin is 1 and Equation is not valid. Otherwise, the equation, valid for any angle must be used. This differential equation can not be easily solved to determine the period of oscillation. However, a simple experiment will show that the period of oscillation for an angle of 90° is only about 15% greater than for an angle of 5°. QUICK QUIZ 15.3 ANSWER

7 A damped oscillator undergoes exponentially decaying oscillatory motion, as in the figure shown here. After 5 seconds of such motion, the amplitude (the quantity Ae -bt/2m ) is a factor of 2 smaller than its initial value at t = 0. After 15 seconds of motion, the amplitude will be a) a factor of 4 smaller than its initial value at t = 0, b) a factor of 6 smaller than its initial value at t = 0, c) a factor of 8 smaller than its initial value at t = 0, or d) impossible to determine without knowing the constants b and m. (end of section 15.6) QUICK QUIZ 15.4

8 (c). The negative exponential function has the property that it will decrease by the same factor over the same period of time. Relating the amplitude after 5 seconds (t 1 ) to the initial amplitude we have, After another 5 seconds we have, So after 15 seconds we get, QUICK QUIZ 15.4 ANSWER

9 Vibration isolation tables are designed to isolate sensitive instruments, for example lasers and microscopes, from vibrations in the floor. The tables behave like a mass and a spring with small damping and act like a forced oscillator with the floor as the driving force. Typical floors vibrate with a frequency of about 20 Hz and this driving frequency is applied to the table. It is desirable to keep the amplitude of vibration of the isolation table to a minimum and the natural frequency of the table is adjusted to be significantly smaller than the driving frequency of the floor. Isolation tables should therefore be designed to have a) a large mass and a small spring constant, b) a small mass and a large spring constant, c) a large mass and a large spring constant, or d) a small mass and a small spring constant. (end of section 15.7) QUICK QUIZ 15.5

10 (a). The equation for the natural frequency, o = (k/m) 1/2, indicates that the spring constant should be small and the mass large to produce a small natural frequency. From examination of Equation 15.36, a large difference in and o, and in addition a large mass m, will reduce the vibration amplitude of the table by a substantial amount. QUICK QUIZ 15.5 ANSWER


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