Question

Two different simple harmonic oscillators have the same natural frequency (f=8.80 Hz) when they are on the surface of the Earth. The first oscillator is a vertical spring and mass, the second is a pendulum. If both systems are moved to the surface of the moon (g=1.67 m/s2), what is the new frequency of the vertical spring and mass?

Answer 1

Answer:

8.80 Hz

Explanation:

The frequency of a loaded spring is given by

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

where k and m are the spring constant and the mass of the load respectively. The values of these do not change because they are internal properties of the components of the system.

Hence, the frequency of the vertical spring mass does not change and is 8.80 Hz.

On the other hand, the frequency of the simple pendulum is affected because it is given by

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

where g and l are acceleration due to gravity and length of the pendulum, respectively. It is thus seen that it depends on g, which changes with location. In fact, the new frequency is given by

$f_2 = 8.80\sqrt{\dfrac{1.67}{9.81}}=3.63 \text{ Hz}$

Answer 2

The new frequency of the vertical spring and mass on moon is 3.63 Hz.

The given parameters;

• natural frequency, f = 8.8 Hz
• acceleration due to gravity on moon, = 1.67 m/s²
• acceleration due to gravity on Earth, = 9.8 m/s²

The frequency of a pendulum at a given length and gravity is calculated as;

$F = \frac{1}{2\pi } \sqrt{\frac{g}{l} } \\\\F = k\sqrt{g} \\\\\frac{F_1}{\sqrt{g_1} } = \frac{F_2 }{\sqrt{g_2} } \\\\F_2 = F_1\sqrt{\frac{g_2}{g_1} } \\\\F_2 = 8.8 \times \sqrt{\frac{1.67}{9.8} } \\\\F_2 = 3.63 \ Hz$

Thus, the new frequency of the vertical spring and mass on moon is 3.63 Hz.

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