Question

At an outdoor market, a bunch of bananas is set on a spring scale to measure the weight. The spring sets the full bunch of bananas into vertical oscillatory motion, which is harmonic with an amplitude 0.21 m. The maximum speed of the bananas is observed to be 1.92 m/s. What is the mass of the bananas? The spring of the scale has a force constant 223 N/m. Answer in units of kg.

Answer 1

2.67kg

Step-by-step explanation:

The maximum velocity,
`v _ {max}` of a body experiencing simple harmonic motion is given by equation (1);

`v_(max)=\omega A............(1)`

where
`\omega` is the angular velocity and A is the amplitude.

The problem describes the oscillation of a loaded spring, and for a loaded spring the angular velocity is given by equation (2);

`\omega=\sqrt{(k)/(m)}.................(2)`

where k is the force constant of the spring and m is the loaded mass.

We can make
`\omega` the subject of formula in equation (1) as follows;

`\omega=(v_(max))/(A).................(3)`

We then combine equations (2) and (3) as follows;

`(v_(max))/(A)=\sqrt{(k)/(m)}.................(4)`

According to the problem, the following are given;

`v_ {max }=1.92m/s\\A=0.21m\\k=223N/m`

We then substitute these values into equation (4) and solve for the unknown mass m as follows;

`(1.92)/(0.21)=\sqrt{(223)/(m)}`

`9.143=\sqrt{(223)/(m)}`

Squaring both sides, we obtain the following;

`9.143^2=(223)/(m)\\9.143^2*m=223\\83.592m=223\\therefore\\m=(223)/(83.592)\\m=2.67kg`