Answer to Question #160914 in Classical Mechanics for Bawe

Question

At an outdoor market, a bunch of bananas attached to the bottom of a vertical spring of force constant 16.0N/m is set into oscillatory motion with amplitude of 20.0cm. It is observed that the maximum speed of the bunch of bananas is 40.0cm/s. What is the weight of the bananas?

Answer 1

The maximum speed of the bunch of bananas undergoing simple harmonic motion can be found as follows:

$$v_{max}=\omega A=\sqrt{\dfrac{k}{m}}A.$$

Then, from this equation, we can find the mass of the bananas:

$$m=\dfrac{kA^2}{v_{max}^2}.$$

Finally, we can find the weight of the bananas:

$$W=mg=\dfrac{kA^2g}{v_{max}^2},$$ $$W=\dfrac{16.0\ \dfrac{N}{m}\cdot(0.2\ m)^2\cdot9.8\ \dfrac{m}{s^2}}{(40\ \dfrac{cm}{s}\cdot\dfrac{1\ m}{100\ cm})^2}=39.2\ N.$$