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Mass-spring system, clock pendulum, submarine sonar echo

1. A mass - spring system in SHM in the horizontal direction. If the mass is 0.25 kg, the spring constant is 12 N/m, and the amplitude is 15 cm, (a) what is the maximum speed of the mass, and (b) where does this occur? (c) What is the speed at half amplitude position?

2. A clock uses a pendulum that is 75 cm long. The clock is accidently broken, and when it is repaired, the length of the pendulum is shortened by 2 mm. Consider the pendulum to be simple pendulum. (a) Will the repaired clock gain or lose time? (b) By how much will the time indicated by the repaired clock differ from the correct time (taken to be the time determined by the original pendulum in 24 h)? (c) If the pendulum were metal, would the surrounding temperature make a difference in timekeeping of the clock? Explain.

3. A submarine on the ocean surface receives a sonar echo indicating an underwater object. The echo comes back at an angle of 20 degrees above the horizontal and the echo took 2.32 s to get back to submarine. What is the object's depth?

Solution Preview

Please refer to the attachment.

1. A mass - spring system in SHM in the horizontal direction. If the mass is 0.25 kg, the spring constant is 12 N/m, and the amplitude is 15 cm, (a) what is the maximum speed of the mass, and (b) where does this occur? (c) What is the speed at half amplitude position?
Solution:

0.25 kg
12 N/m

a) The amplitude (A) is given as 15 cm. At the maximum amplitude the mass comes to rest momentarily, hence its kinetic energy is zero. Whole of the mechanical energy is converted into potential energy of the spring. When the mass is passing through its mean position (spring un-deformed) the PE of the spring is zero and whole of the mechanical energy of the system is in the form of KE of the mass. Hence, its speed at the mean position is maximum.

½ mv2 = ½ ...

Solution Summary

A mass-spring system, clock pendulum and submarine sonar echo's are examined.

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