a.
- i. the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s
- ii. the angular velocity when the two springs are in parallel is 7.07 rad/s
b.
- i. the speed of the block of mass when the springs are connected in series is 11.2 A m/s
- ii. the angular velocity when the two springs are in series is 11.2 rad/s
Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is k' = k + k
= 2k
= 2 × 250 N/m
= 500 N/m
Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA
Also, 1/2k'A² = 1/2k'x² + 1/2Mv² where
- k' = equivalent spring constant in parallel = 500 N/m,
- A = maximum displacement of spring,
- x = equilibrium position = 0 m,
- M = mass of block = 10 kg and
- v = speed of block at equilibrium position
Making v subject of the formula, we have
v = √[k'(A² - x²)/M]
Substituting the values of the variables into the equation, we have
v = √[k'(A² - x²)/M]
v = √[500 N/m(A² - (0)²)/10]
v = √[50 N/m(A² - 0)]
v = [√50]A m/s
v = [5√2] A m/s
v = 7.07 A m/s
So, the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s
<h3>ii. The angular velocity of mass when the springs are in parallel</h3>
Since velocity of spring v = ω√(A² - x²) where
- ω = angular velocity of spring,
- A = maximum displacement of spring and
- x = equilbrium position of spring = 0 m
Making ω subject of the formula, we have
ω = v/√(A² - x²)
Since v = 7.07 A m/s
Substituting the values of the other variables into the equation, we have
ω = v/√(A² - x²)
ω = 7.07 A m/s/√(A² - 0²)
ω = 7.07 A m/s/√(A² - 0)
ω = 7.07 A m/s/√A²
ω = 7.07 A m/s/A m
ω = 7.07 rad/s
So, the angular velocity when the two springs are in parallel is 7.07 rad/s
<h3>b. </h3><h3>i. How to calculate the velocity of the mass when the springs are connected in series?</h3>Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is 1/k" = 1/k + 1/k
= 2/k
⇒ k" = k/2
k" = 250 N/m ÷ 2
= 125 N/m
Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA
Also, 1/2k"A² = 1/2k"x² + 1/2Mv'² where
- k" = equivalent spring constant in series = 125 N/m,
- A = maximum displacement of spring,
- x = equilibrium position = 0 m,
- M = mass of block = 10 kg and
- v' = speed of block at equilibrium position
Making v subject of the formula, we have
v = √[k"(A² - x²)/M]
Substituting the values of the variables into the equation, we have
v = √[k"(A² - x²)/M]
v = √[125 N/m(A² - (0)²)/10]
v = √[125 N/m(A² - 0)]
v = [√125]A m/s
v = [5√5] A m/s
v = 11.2 A m/s
So, the speed of the block of mass when the springs are connected in series is 11.2 A m/s
<h3>ii. The angular velocity of the mass when the springs are in series</h3>
Since velocity of spring v = ω√(A² - x²) where
- ω = angular velocity of spring,
- A = maximum displacement of spring and
- x = equilbrium position of spring = 0 m
Making ω subject of the formula, we have
ω = v/√(A² - x²)
Since v = 11.2 A m/s
Substituting the values of the other variables into the equation, we have
ω = v/√(A² - x²)
ω = 11.2 A m/s/√(A² - 0²)
ω = 11.2 A m/s/√(A² - 0)
ω = 11.2 A m/s/√A²
ω = 11.2 A m/s/A m
ω = 11.2 rad/s
So, the angular velocity when the two springs are in series is 11.2 rad/s
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