Answer :
Answer:
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
Answer:
The mass of the block is 496 g
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
[tex]E = \frac{1}{2} k(\frac{A}{2} )^{2} + \frac{1}{2} mv^{2} = \frac{E}{4} + \frac{1}{2} mv^{2}[/tex]
Where:
m = Mass of the block
v = Velocity of block at the instant (Halfway between its equilibrium position and the endpoint)
Therefore,
[tex]\frac{3}{4}E = \frac{1}{2} mv^{2}[/tex] or
m = [tex]\frac{3}{2v^2}E[/tex] = [tex]\frac{3}{2\times 0.313^2} \times 3.24 \times 10^{-2}[/tex]= 0.496 kg = 496 g.