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Figure 1 Figure 2 1. (7 points, suggested time 13 minutes) A cart on a hori...

Feb 20, 2024

Figure 1 Figure 2 1. (7 points, suggested time 13 minutes) A cart on a horizontal surface is attached to a spring. The other end of the spring is attached to a wall. The cart is initially held at rest, as shown in Figure 1. When the cart is released, the system consisting of the cart and spring oscillates between the positions

x=+L

and

x=-L

. Figure 2 shows the kinetic energy of the cart-spring system as a function of the system's potential energy. Frictional forces are negligible. (a) On the graph of kinetic energy

K

versus potential energy

U

shown in Figure 2, the values for the

x

-intercept and

y

-intercept are the same. Briefly explain why this is true, using physics principles. Figure 3 When the cart is at

+L

and momentarily at rest, a block is dropped onto the cart, as shown in Figure 3 . The block sticks to the cart, and the block-cart-spring system continues to oscillate between

-\boldsymbol{L}

and

+\boldsymbol{L}

. The masses of the cart and the block are

m_{0}

and

3 m_{0}

, respectively. (b) The frequency of oscillation before the block is dropped onto the cart is

f_{1}

. The frequency of oscillation after the block is dropped onto the cart is

f_{2}

. Calculate the numerical value of the ratio

\frac{f_{2}}{f_{1}}

Solution by Steps

step 1

To solve the equation

(\frac{f2}{f1})^2 = \frac{m0}{4m0}

, we first simplify the right side of the equation

step 2

Simplifying the fraction

\frac{m0}{4m0}

gives us

\frac{1}{4}

because

m0

cancels out in the numerator and denominator

step 3

Now we have

(\frac{f2}{f1})^2 = \frac{1}{4}

. To find

\frac{f2}{f1}

, we take the square root of both sides

step 4

Taking the square root of both sides gives us

\frac{f2}{f1} = \pm \frac{1}{2}

. However, since frequency cannot be negative, we discard the negative solution

Answer

\frac{f2}{f1} = \frac{1}{2}

Key Concept

Solving a ratio of squares for the original ratio

Explanation

The ratio of the squares of two frequencies is equal to the ratio of their squares, and when taking the square root, we consider only the positive solution since frequency cannot be negative.