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 0.15 m. The maximum speed of the bananas is observed to be 2.34 m/s. What is the mass of the bananas

Complete question is;

At an outdoor market, a bunch of bananas is set on a spring scale to measure the weight. The spring of force constant 16.0 N/m sets the full bunch of bananas into vertical oscillatory motion, which is harmonic with an amplitude 0.15 m. The maximum speed of the bananas is observed to be 2.34 m/s. What is the mass of the bananas

Answer:

Mass of banana = 0.0657 kg

Explanation:

We are given;

Amplitude; A = 0.15m

Spring constant; k = 16 N/m

Maximum velocity; v_max = 2.34 m/s

Now, In oscillatory motion, maximum velocity is given by;

V_max = Aω

Where A is amplitude and ω is angular velocity.

Making ω the subject, we have;

ω = V_max/A

ω = 2.34/0.15

ω = 15.6 rad/s

Now, For spring, angular velocity is given as;

ω = √ (k/m)

Where k is force constant and m is mass.

Thus, plugging in the relevant values, we have;

15.6 = √ (16/m)

15.6² = 16/m

m = 16/15.6²

m ≈ 0.0657 kg