Question

when mass of 40kg attached to vertically hanging spring it extends by 0.4cm find1
- force constant of spring
-extension when 100g weight is attached to
- time period of oscillation of 100g weight on it
- time period and force constant if spring is cut in 3 equal part and 100g weight is made to oscillate on one part

Answer 1

Answer on Question 57319, Physics, Mechanics | Relativity

Question:

When a mass of $40\,g$ is attached to a vertically hanging spring it extends by $0.4\,cm$. Find:

1) Force constant of the spring.

2) The extension when $100\,g$ weight is attached to it.

3) The time period of oscillation of $100\,g$ weight on it.

4) The time period and force constant if the spring is cut in three equal parts and $100\,g$ weight is made to oscillate on one part.

Solution:

1) We can find the force constant of the spring from the Hooke’s law:

$$F = kx,$$

here, $F$ is the force acting on the spring, $k$ is the spring constant (force constant of the spring), $x$ is the extension of the spring.

From the other hand, $F = mg$, and substituting it into the first equation we get:

$$mg = kx,$$ $$k = \frac{mg}{x} = \frac{0.04kg \cdot 9.8\,\frac{m}{s^2}}{0.004m} = 98\,\frac{N}{m}.$$

2) As we know the force constant of the spring, we can find from the previous formula the extension of the spring when $100\,g$ weight is attached to it:

$$x = \frac{mg}{k} = \frac{0.1kg \cdot 9.8\,\frac{m}{s^2}}{98\,\frac{N}{m}} = 0.01m.$$

3) We can find the time period of oscillation of $100\,g$ weight from the formula:

$$T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.1kg}{98\,\frac{N}{m}}} = 0.2s.$$

4) Let's assume, that $F = k_{big}x_{big}$ is the force that acts on the spring. Then, the spring is divided to three equal small springs. The force that acts on each of these small springs is the same (equal to the force that acts on the whole spring) and must be equal to $F = k_{small}x_{small}$. Because $x_{small} = \frac{x_{big}}{3}$ we get:

$$k_{big}x_{big} = k_{small}x_{small},$$ $$k_{big}x_{big} = k_{small}\frac{x_{big}}{3},$$ $$k_{big} = \frac{k_{small}}{3},$$ $$k_{small} = 3k_{big} = 3 \cdot 98\frac{N}{m} = 294\frac{N}{m}.$$

Then, we can find the time period:

$$T_{new} = 2\pi \sqrt{\frac{m}{3k_{big}}} = 2\pi \sqrt{\frac{0.1kg}{3 \cdot 98\frac{N}{m}}} = 0.11s.$$

Answer:

1) $k = 98\frac{N}{m}$.

2) $x = 0.01m$.

3) $T = 0.2s$.

4) $k_{small} = 294\frac{N}{m}, T_{new} = 0.11s$.

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