If the pendulum of a clock goes back and forth 35 times in 1 minute with an amplitude of 2 meters, calculate the maximum speed it reaches.

A. [tex]V_{\text{max}} = 4.43 \, \text{m/s}[/tex]
B. [tex]V_{\text{max}} = 8.86 \, \text{m/s}[/tex]
C. [tex]V_{\text{max}} = 1.57 \, \text{m/s}[/tex]
D. [tex]V_{\text{max}} = 2.79 \, \text{m/s}[/tex]

Through the use of the formula Umax = Aw, we calculate the maximum speed of a pendulum with a given amplitude and frequency. However, the calculated speed does not match any of the options provided in the question.

Explanation:

The problem at hand is requiring us to calculate the maximum speed of a pendulum. For physics problems like this, it's helpful to utilize the formula of the maximum speed of a harmonic oscillator, which is the amplitude (A) times the angular frequency (w). The formula is expressed as Umax = Aw.

In this particular case, the given amplitude (A) is 2 meters, and the angular frequency (w) can be derived from the pendulum's oscillations. Since the pendulum made 35 oscillations in one minute, this presents a frequency of 35/60 = 0.583 Hz. To convert this to angular frequency, we multiply by 2π, obtaining w = 2π*0.583 = 3.66 rad/sec.

Finally, using the formula Umax = Aw, we plug in our amplitude and angular frequency values, getting Umax = 2*3.66 = 7.32 m/s. However, none of the options in the question match this answer. Therefore, there might be an error in the provided options or in the premises of the question.

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