Chapter 4: Q53P (page 168)

Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude $1.7 c m$ . You observe that it takes $2.1 s$ to make one round-trip. You construct another vertical oscillator with a mass 6 times as heavy and a spring 10 times as stiff. You take it to a planet where $g_{p l a n e t} = 6.8 N / k g .$ You start oscillating with amplitude $3.3 c m$ . How long does it take for the mass to make one round-trip?

Expert verified

The period $T' = 1.627 s$ it takes for the mass to make one round trip.

Step by step solution

Given data can be listed below,

Acceleration due to the gravity of the planet, $g_{p l a n e t} = 6.8 N / k g$

Expression for angular frequency of the spring-mass system is given by

$ω = \sqrt{\frac{k}{m}}$

Our construction spring-mass system and the angular frequency are given by

$ω' = \sqrt{\frac{k'}{m'}} = \sqrt{\frac{10 k}{6 m}} = 1.29 \sqrt{\frac{k}{m}} = 1.29 ω$

The expression for the period of the spring-mass system is given by

$ω = 2 π f = \frac{2 π}{T} T = \frac{2 π}{ω}$

The period of the oscillator does not depend on gravity as the measurement of horizontal spring.

From the period of our construction spring-mass system is given by

$T' = \frac{2 π}{ω'}$

Now from the above equation, we can write as

$T ω = 2 π T' = T \frac{ω}{1.29 ω} = 2.1 \times \frac{1}{1.29} = 1.627 s$

Thus, the period $T' = 1.627 s$ it takes for the mass to make one round trip.

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