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3 years ago

7

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The sprin

gs of a certain car are adjusted so that the oscillations have a frequency of 4.42 Hz. (a) What is the spring constant of each spring if the mass of the car is 1450 kg and the mass is evenly distributed over the spring?

1 answer:

iogann1982 [59]3 years ago

5 0

Answer:

Therefore, the spring constant of each spring = 1.6 × 10⁻⁶ kg/s².

Explanation:

The period (T) of a spring in oscillation = 2π √(m/k)............. equation 1

Where m = mass acting on the spring (kg), k = spring constant of the spring (kg/s²).

Making k the subject of equation 1

k = T²/(4π²×m) .......................... equation 2

From the question, F = 4.42 Hz,

since T = 1/F

then, T = 1/F = 1/4.42 =0.226 s, π = 3.143

since the weight of the mass is evenly distributed over the four identical spring, Hence

m = 1450/4 = 362.5 kg

Substituting these values into equation 2

k = 0.226/{(4×3.143²)362.5}

k = 0.226/(14323.751)

k = 0.0000016 kg/s²

k = 1.6 × 10⁻⁶ kg/s².

Therefore, the spring constant of each spring = 1.6 × 10⁻⁶ kg/s².

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A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant

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<h2>Answer: 13.61 N/m</h2>

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force $F$ applied to it, <u>as long as the spring is not permanently deformed</u>:

$F=k (x-x_{o})$ (1)

Where:

$k$ is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.

$x_{o}$ is the length of the spring without applying force.

$x$ is the length of the spring with the force applied.

According to this, we have a spring where only the force due gravity is applied.

In other words, the force applied is the weigth $W$ of the block:

$W=m.g$ (2)

Where $m=250g=0.25kg$ is the mass of the block and $g=9.8\frac{m}{s^{2}}$ is the gravity acceleration.

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$W=2.45N$ (4)

Knowing the force applied $W$ and $x=18cm=0.18m$ and $x_{o}=0$ , we can substitute the values in equation (1) and find $k$ :

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$2.45N=k (0.18m-0m)$ (6)

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