Question

A block of unknown mass is attached to a spring of spring constant 6.1 N/m and undergoes simple harmonic motion with an amplitude of 10.3 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be 31.3 cm/s. Calculate the mass of the block. Answer in units of kg

Answer 1

The mass of the block is 496 g

Step-by-step explanation:

Here we have by the principle of conservation of energy;

Energy in spring, E = 0.5·k·A²

Where:

k = Spring constant = 6.1 N/m

A = Amplitude of motion = 10.3 cm = 0.103 m

E = 0.5×6.1×0.103² = ‭0.03235745 J= 3.24 × 10⁻² J

At half way, we have

`E = (1)/(2) k((A)/(2) )^(2) + (1)/(2) mv^(2) = (E)/(4) + (1)/(2) mv^(2)`

Where:

m = Mass of the block

v = Velocity of block at the instant (Halfway between its equilibrium position and the endpoint)

Therefore,

`(3)/(4)E = (1)/(2) mv^(2)` or

m =
`(3)/(2v^2)E` =
`(3)/(2* 0.313^2) * 3.24 * 10^(-2)`= 0.496 kg = 496 g.

Answer 2

Step-by-step explanation:

Given that, .

Spring constant.

K = 6.1 N/m

Amplitude of oscillation

A = 10.3cm = 0.103m

Half wave, it speed is

V = 31.3cm/s = 0.313m/s

Mass of block

M =?

The is half of the distance between the equilibrium position and endpoint

Then,

x = A/2

Where,

A is the amplitude

x is the position of the block

x = A / 2 = 0.103/2

x = 0.0515m

The velocity if the block at any given position is given as

v = ω√(A²—x²)

Then,

ω = v / √(A²—x²)

Where

ω is angular frequency

v Is the velocity of the block

The force constant is given as

k = mω²

Where,

K is spring constant

ω is angular frequency

Substitute ω into k

Then,

k = m (v / √(A²—x²))²

k = mv² / (A²—x²)

Make m subject of formula since we want to find m

m = k(A²—x²) / v²

m = 6.1 (0.103²—0.0515²) / 0.313²

m = 6.1 × 7.96 × 10^-3 / 0.313²

m = 0.495 kg

The mass of the block is approximately 0.5kg