Question

A mass on a spring vibrates in simple harmonic motion at a frequency of 3.26 Hz and an amplitude of 5.76 cm. If the mass of the object of 0.218 kg, what is the spring constant

Answer 1

91.48N/m

Step-by-step explanation:

In a spring-mass system undergoing a simple harmonic motion, the inverse of the frequency f, of oscillation is proportional to the square root of the mass m, and inversely proportional to the square root of the spring constant, k. This can be expressed mathematically as follows;

`(1)/(f)` =
`2\pi\sqrt{(m)/(k) }` -----------(i)

From the question;

f = 3.26 Hz

m = 0.218kg

Substitute these values into equation (i) as follows;

`(1)/(3.26)` =
`2\pi\sqrt{(0.218)/(k) }` [Square both sides]

(
`(1)/(3.26)`)² = (
`2\pi`)²(
`(0.218)/(k)`)

(
`(1)/(10.6276)`) =
`4\pi`²(
`(0.218)/(k)`) [Take
`\pi` to be 3.142]

(
`(1)/(10.6276)`) =
`4(3.142)`²(
`(0.218)/(k)`)

(
`(1)/(10.6276)`) =
`39.488`(
`(0.218)/(k)`)

(
`(1)/(10.6276)`) = (
`(8.608)/(k)`) [Switch sides]

(
`(8.608)/(k)`) = (
`(1)/(10.6276)`) [Re-arrange]

(
`(k)/(8.608)`) = (
`(10.6276)/(1)`) [Cross-multiply]

k = 8.608 x 10.6276

k = 91.48N/m

Therefore, the spring constant of the spring is 91.48N/m