High School

A block of unknown mass is attached to a spring with a spring constant of 7.4 N/m and undergoes simple harmonic motion with an amplitude of 5.5 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be 38.8 cm/s. Calculate the mass of the block. Answer in units of kg.

Answer :

Final answer:

The mass of the block is 0.800 kg.

Explanation:

To calculate the mass of the block, we can use the given information and the formulas for simple harmonic motion.

First, let's calculate the period of oscillation using the formula T = 2π√(m/k), where T is the period, m is the mass of the block, and k is the spring constant.

Since the block is halfway between its equilibrium position and the endpoint, its displacement is equal to half the amplitude, which is 5.5 cm / 2 = 2.75 cm.

The speed of the block at this point is given as 38.8 cm/s.

Using the formula v = ωA, where v is the speed, ω is the angular frequency, and A is the amplitude, we can calculate the angular frequency:

ω = v / A = 38.8 cm/s / 5.5 cm = 7.05 rad/s.

Now, we can calculate the period using the formula T = 2π/ω:

T = 2π / 7.05 rad/s = 0.895 s.

Finally, we can use the period and the spring constant to calculate the mass of the block:

T = 2π√(m/k)

0.895 s = 2π√(m/7.4 N/m)

Squaring both sides of the equation:

0.800 m^2/s^2 = 4π^2(m/7.4 N/m)

Simplifying the equation:

0.800 m^2/s^2 = (4π^2/7.4 N/m) * m

0.800 m^2/s^2 = (1.698 m^2/s^2) * m

Dividing both sides of the equation by 1.698 m^2/s^2:

0.800 = m

Therefore, the mass of the block is 0.800 kg.

Learn more about calculating the mass of a block attached to a spring here:

https://brainly.com/question/30266126

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Final answer:

The mass of the block is 0.800 kg.

Explanation:

To calculate the mass of the block, we can use the given information and the formulas for simple harmonic motion.

First, let's calculate the period of oscillation using the formula T = 2π√(m/k), where T is the period, m is the mass of the block, and k is the spring constant.

Since the block is halfway between its equilibrium position and the endpoint, its displacement is equal to half the amplitude, which is 5.5 cm / 2 = 2.75 cm.

The speed of the block at this point is given as 38.8 cm/s.

Using the formula v = ωA, where v is the speed, ω is the angular frequency, and A is the amplitude, we can calculate the angular frequency:

ω = v / A = 38.8 cm/s / 5.5 cm = 7.05 rad/s.

Now, we can calculate the period using the formula T = 2π/ω:

T = 2π / 7.05 rad/s = 0.895 s.

Finally, we can use the period and the spring constant to calculate the mass of the block:

T = 2π√(m/k)

0.895 s = 2π√(m/7.4 N/m)

Squaring both sides of the equation:

0.800 m^2/s^2 = 4π^2(m/7.4 N/m)

Simplifying the equation:

0.800 m^2/s^2 = (4π^2/7.4 N/m) * m

0.800 m^2/s^2 = (1.698 m^2/s^2) * m

Dividing both sides of the equation by 1.698 m^2/s^2:

0.800 = m

Therefore, the mass of the block is 0.800 kg.

Learn more about calculating the mass of a block attached to a spring here:

https://brainly.com/question/30266126

#SPJ14