Physics, 30.06.2019 02:30 tsdean21

Avertical spring (spring constant =160 n/m) is mounted on the floor. a 0.340-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. the block is not attached to the spring. (a) obtain the frequency (in hz) of the motion. (b) determine the amplitude at which the block will lose contact with the spring.

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Answer from: bm42400

a. The frequency (in Hz) of the motion is equal to 3.45 Hertz.

b. The amplitude at which the block will lose contact with the spring is 0.021 meters.

Given the following data:

Spring constant =160 N/mMass of block = 0.340 Kg

a. To find the frequency (in Hz) of the motion:

Since the spring initiates simple harmonic motion, we would determine its angular frequency.

Mathematically, angular frequency of a spring is given by the formula:

$\omega = \sqrt{\frac{k}{m} }$

Where:

k is the spring constant.m is the mass of the spring.

Substituting the given parameters into the formula, we have;

$\omega = \sqrt{\frac{160}{0.340} }\\\\\omega = \sqrt{470.59}\\\\\omega = 21.69 \;rad/s$

Now, we can find the frequency of the motion by using the formula:

$F = \frac{\omega}{2\pi} \\\\F = \frac{21.69}{2 \times 3.142} \\\\F = \frac{21.69}{6.284}$

Frequency, F = 3.45 Hz

b. To determine the amplitude at which the block will lose contact with the spring:

$Force = kA\\\\mg = kA\\\\A = \frac{mg}{k}$

Where:

A is the amplitude.k is the spring constant.m is the mass of the spring.

Substituting the parameters into the formula, we have;

$A = \frac{0.340 \times 9.8}{160} \\\\A = \frac{3.332}{160}$

Amplitude, A = 0.021 meters

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