Vibration and Wave Fundamentals
1. A machine gun fires 10 rounds per second. The speed of the bullets is 300 m/s.
(A.) What is the distance in the air between the flying bullets? ____
ANSWER=30
(B.)What happens to the distance between the bullets if the rate of fire is increased? ______
ANSWER=Decreases
2. Consider a wave generator that produces 10 pulses per second. The speed of the waves is 300 cm/s.
(A.) What is the wavelength of the waves? ______
ANSWER=30
(B.) What happens to the wavelength if the frequency of pulses is increased? ______
ANSWER=Increases
3. (A.) A bird is watching waves. If the portion of a wave between 2 crests passes the pole each second, what is the speed of the wave? (1 m between crests)____
ANSWER=1m/s
(B.) What is its period? ____
ANSWER=1
4. (A.) If the distance between crests in the above question were 1.5 meters apart, and 2 crests pass the pole each second, what would be the speed of the wave?_________
ANSWER=1.5m/s
(B.) What would be its period? ______
ANSWER=1
Thanks so much for the help!!
1. (A.) To find the distance between the flying bullets, we can use the formula: distance = speed × time. Since the machine gun fires 10 rounds per second and the speed of the bullets is 300 m/s, the distance between the bullets can be calculated as follows:
Distance = 300 m/s × (1 s / 10 rounds) = 30 m.
Therefore, the distance between the flying bullets is 30 meters.
(B.) If the rate of fire is increased, meaning more bullets are fired per second, the distance between the bullets will decrease. This is because the time interval between each bullet decreases, resulting in a shorter distance between them.
2. (A.) The wavelength of waves can be calculated using the formula: wavelength = speed / frequency. Given that the wave generator produces 10 pulses per second and the speed of the waves is 300 cm/s, we can find the wavelength as follows:
Wavelength = 300 cm/s / 10 pulses/s = 30 cm.
Therefore, the wavelength of the waves is 30 cm.
(B.) If the frequency of pulses is increased, meaning more pulses are produced per second, then the wavelength of the waves will increase. This is because the speed of the waves remains constant, but with a higher frequency, the waves will have more peaks and troughs within a given distance, resulting in a longer wavelength.
3. (A.) The speed of a wave can be calculated using the formula: speed = wavelength × frequency. In this case, the portion of the wave between 2 crests passes the pole each second, and the distance between crests is 1 m. So, the speed of the wave would be:
Speed = 1 m/crest × 2 crests/s = 2 m/s.
Hence, the speed of the wave is 2 m/s.
(B.) The period of a wave is the time it takes for one complete cycle, which is the reciprocal of the frequency. As mentioned earlier, the portion of the wave between 2 crests passes the pole each second, so the frequency is 1 cycle per second. Therefore, the period of the wave would be:
Period = 1 s / 1 cycle/s = 1 s.
Thus, the period of the wave is 1 second.
4. (A.) Similar to question 3, the speed of the wave can be calculated using the formula: speed = wavelength × frequency. Given that the distance between crests is now 1.5 meters and 2 crests pass the pole each second, the speed of the wave is:
Speed = 1.5 m/crest × 2 crests/s = 3 m/s.
Therefore, the speed of the wave is 3 m/s.
(B.) Since the frequency remains the same at 2 crests per second, the period of the wave would be:
Period = 1 s / 2 crests/s = 0.5 s (or 1/2 s).
Hence, the period of the wave is 0.5 seconds.