Question

1.A 0.321-kg mass is attached to a spring with a force constant of 53.7 N/m.

If the mass is displaced 0.240 m from equilibrium and released, what is its speed when it is 0.138 m from equilibrium?

v=____m/s

2.A Cricket Thermometer, by Jiminy Insects are ectothermic, which means their body temperature is largely determined by the temperature of their surroundings. This can have a number of interesting consequences. For example, the wing coloration in some butterfly species is determined by the ambient temperature, as is the body color of several species of dragonfly. In addition, the wing beat frequency of beetles taking flight varies with temperature due to changes in the resonant frequency of their thorax.
The origin of such temperature effects can be traced back to the fact that molecules have higher speeds and greater energy as temperature is increased (see Chapters 16 and 17 in the textbook). Thus, for example, molecules that collide and react as part of the metabolic process will do so more rapidly when the reactions are occurring at a higher temperature. As a result, development rates, heart rates, wing beats, and other processes all occur more rapidly.
One of the most interesting thermal effects is the temperature dependence of chirp rate in certain insects. This behavior has been observed in cone-headed grasshoppers, as well as several types of cricket. A particularly accurate connection between chirp rate and temperature is found in the snowy tree cricket (Oecanthus fultoni Walker), which chirps at a rate that follows the expression N=T−39, where N is the number of chirps in 13 seconds, and T is the numerical value of the temperature in degrees Fahrenheit. This formula, which is known as Dolbear's law, is plotted in the figure(Figure 1) (green line) along with data points (blue dots) for the snowy tree cricket.

If the temperature is increased by 10 degrees Fahrenheit, how many additional chirps are heard in a 13-s interval?

Answer 1

53.7 1 m = 0.321 kg, k = 53.7 N/m T= 274 m 2x 3014 0:32] T= 0.486 sec A = 0.24 m x = 0.138m v=w /A²x2 wo 25 J624)2-(0-138)2 u : 2x 3.14 0.486 v 0.038556 =2.54 m/s