Question

problem 7:
a space traveler of mass 80 kg lands on a new planet. the traveler takes out a simple pendulum of length 0.2 m and measures the frequency of oscillations to be 1.77 hz. what is the weight in newtons of the traveler? is the traveler heavier or lighter than on earth? explain.

Explanation:

Step1: Find the formula for the frequency of a simple pendulum

The formula for the frequency \( f \) of a simple pendulum is \( f=\frac{1}{2\pi}\sqrt{\frac{g}{L}} \), where \( g \) is the acceleration due to gravity and \( L \) is the length of the pendulum.

Step2: Solve the formula for \( g \)

First, square both sides of the formula \( f=\frac{1}{2\pi}\sqrt{\frac{g}{L}} \) to get \( f^{2}=\frac{1}{4\pi^{2}}\frac{g}{L} \). Then, solve for \( g \):

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Substitute \( f = 1.77\space Hz\) and \( L=0.2\space m\) into the formula:

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Step3: Calculate the weight \( W \) of the traveler

The weight formula is \( W = mg \). Given \( m = 80\space kg\) and \( g = 24.71\space m/s^{2}\), then \( W=80\times24.71 = 1976.8\space N\)

On Earth, \( g_{Earth}=9.8\space m/s^{2}\), and the weight on Earth \( W_{Earth}=mg_{Earth}=80\times9.8 = 784\space N\)

Since \( 1976.8>784\), the traveler is heavier on the new planet.

Answer:

The weight of the traveler is \( 1976.8\space N\). The traveler is heavier than on Earth because the acceleration due to gravity \( g\) on the new planet (\(24.71\space m/s^{2}\)) is greater than the acceleration due to gravity on Earth (\(9.8\space m/s^{2}\)), and weight \(W = mg\) (mass \(m\) is constant).