A vibration platform oscillates up and down with a fixed amplitude of 9.7 cm and a controlled frequency that can be varied. If a small rock of unknown mass is placed on the platform, at what frequency will the rock just begin to leave the surface so that it starts to clatter?

Answers

Answer 1

Answer:

1.6 Hz

Explanation:

In simple harmonic motion, acceleration is given by the formula;

a = Aω²

Where;

A is amplitude

ω is angular velocity.

We are given A = 9.7 cm = 0.097 m

We know that acceleration due to gravity is 9.8 m/s².

Thus;

9.8 = 0.097ω²

ω² = 9.8/0.097

ω = √101.031

ω = 10.05 rad/s

Formula for frequency is;

f = ω/2π

Thus;

f = 10.05/2π

f = 1.6 Hz


Related Questions

Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water
at 6◦ C. How many grams of ice will melt? The heat of fusion of water is 333 kJ/kg and the
specific heat is 4190 J/kg · K.Immersive Reader

Answers

Answer:

7.55 g

Explanation:

Given that:

Heat of fusion = 333kj/kg

Heat capacity, c = 4190 j/kg /k

The Number of grams of ice that will melt can be represented as y:

Number of grams of ice that will melt * heat of fusion = specific heat capacity * temperature change

y * 333 * 10^3 J = (4190) * (6 - 0)

333000y = 25140

y = 25140 / 333000

y = 0.0754954 kg

y = 0.0754954 * 100

y = 7.549 g

Hence, Number of grams of ice that will melt = 7.55 g

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