Question

n this example we will look at the periodic motion of an ultrasonic transducer, a kind of ultrasonic loudspeaker used for medical diagnosis. The transducer is oscillating at a frequency of 6.7 MHz (6.7×106Hz). How much time does each oscillation take, and what is the angular frequency?

Answer 1

The time for each oscillation is approximately 1.49 x 10^-7 seconds, and the angular frequency is approximately 4.21 x 10^7 rad/s.

Step-by-step explanation:

To find the time for each oscillation, we can use the formula T = 1/f, where T represents the period and f represents the frequency. In this case, the frequency is given as 6.7 MHz, which is equivalent to 6.7 x 10^6 Hz. So, substituting the value of the frequency into the formula, we get: T = 1 / (6.7 x 10^6) = 1.49 x 10^-7 seconds. Therefore, each oscillation takes approximately 1.49 x 10^-7 seconds.

The angular frequency can be calculated using the formula ω = 2πf, where ω represents the angular frequency and f represents the frequency. Substituting the given value of the frequency into the formula, we get: ω = 2π x 6.7 x 10^6 = 4.21 x 10^7 rad/s. Therefore, the angular frequency is approximately 4.21 x 10^7 rad/s.

Answer 2

Step-by-step explanation:

frequency, f = 6.7 MHz = 6.7 x 10^6 Hz

Time period is defined as the time taken by the oscillating body to complete one oscillation.

the formula for the time period is

T = 1/ f = 1 / (6.7 x 10^6) = 1.5 x 10^-7 second

The angular frequency is given by

ω = 2 π f = 2 x 3.14 x 6.7 x 10^6

ω = 4.2 x 10^7 rad/s