With what frequency would you have to push a child on a swing that had supporting chains that were 2.3 m long? Express your answer in Hz to 3 significant digits.

T = 2pi sqrt(L/g)

f = 1/T

Well, if we want to calculate the frequency of pushing a child on a swing, we'll have to take into account a few factors. First, we need to determine the time it takes for a complete swing back and forth, aka the period. Let's assume the child swings back and forth in a regular manner.

To do that, we can use the formula for the period of a pendulum, which happens to be almost identical to a swing like this:

T = 2π √(L/g),

where T is the period, L is the length of the chains (2.3m), and g is the acceleration due to gravity (approximately 9.81 m/s²).

Plugging in the values, we get:

T = 2π √(2.3 / 9.81) ≈ 2.030 seconds.

Now that we know the period, we can calculate the frequency:

f = 1 / T,

f ≈ 1 / 2.030 ≈ 0.493 Hz.

So, the frequency at which you would have to push the child on the swing would be approximately 0.493 Hz. Although, I must say, you might want to consider investing in an automatic swing pushing machine for that level of precision!

To determine the frequency required to push a child on a swing, we can use the formula for the period of a simple pendulum:

T = 2π√(L/g)

where:
T = period (time for one complete swing)
L = length of the pendulum (in this case, the chains) = 2.3 m
g = acceleration due to gravity = 9.8 m/s²

Let's plug in the values:

T = 2π√(2.3/9.8)

T ≈ 2π√0.2347

T ≈ 2π × 0.4842

T ≈ 3.041

The frequency (f) is the reciprocal of the period:

f = 1/T

f ≈ 1/3.041

f ≈ 0.328 Hz

Therefore, to three significant digits, you would have to push the child on the swing with a frequency of approximately 0.328 Hz.

To determine the frequency at which you need to push a child on a swing, we can use the formula for the period of a pendulum. The period of a pendulum is given by the equation:

T = 2π √(L / g)

where T is the period of the pendulum, L is the length of the supporting chains, and g is the acceleration due to gravity (approximately 9.8 m/s²).

In this case, the length of the supporting chains is given as 2.3 m. Plugging this value into the equation, we have:

T = 2π √(2.3 / 9.8)

Simplifying, we get:

T ≈ 2π √(0.2347)

Calculating the square root, we have:

T ≈ 2π * 0.484

Finally, multiplying by 2π, we get:

T ≈ 3.04 seconds

Since frequency is the reciprocal of the period, the frequency can be calculated as:

f = 1 / T ≈ 1 / 3.04 ≈ 0.328 Hz

Therefore, the frequency at which you would have to push a child on a swing with supporting chains that are 2.3 m long is approximately 0.328 Hz.