Question

The formula for the speed of transverse waves on a spring is the same as for a string. (a) A spring is stretched to a length much greater than its relaxed length. Explain why the tension in the spring is approximately proportional to the length. (b) A wave takes 4.00 s to travel from one end of such a spring to the other. Then the length is increased \(10.0 \% .\) Now how long does a wave take to travel the length of the spring? [Hint: Is the mass per unit length constant?]

Short answer

Answer: The tension in a spring is approximately proportional to its length when stretched because of Hooke's Law, which states that the spring force is proportional to the displacement from the equilibrium position. After the length of the spring is increased by 10%, the wave takes 4.4 seconds to travel from one end to another.

Step-by-step solution

Understand Hooke's Law and spring force

According to Hooke's Law, the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position: \(F = -kx\), where \(F\) is the spring force, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. When a spring is stretched to a length much greater than its relaxed length, the tension in the spring becomes approximately equal to the spring force (\(T \approx F\)).

Apply Hooke's Law to spring tension

Since the spring force is proportional to the displacement from the equilibrium position (\(F \propto x\)), we can also say that the tension in the spring is approximately proportional to the displacement (\(T \approx F \propto x\)). Thus, when the spring is stretched to a length much greater than its relaxed length, the tension in the spring is approximately proportional to the length. #b) A wave takes 4.00 s to travel from one end of such a spring to the other. Then the length is increased \(10.0 \% .\) Now how long does a wave take to travel the length of the spring?#

Use given information

From the problem, we know that the wave on a spring takes \(4.00\,\text{s}\) to travel from one end to the other. We are also given that the length is increased by \(10.0 \%\) or \(1.10 L\), where \(L\) is the initial length of the spring.

Apply wave speed equation

The wave speed equation is given by \(v = \frac{d}{t}\), where \(v\) is the wave speed, \(d\) is the distance traveled, and \(t\) is the time taken. We can write this equation for the initial length (\(L\)) and the increased length (\(1.10 L\)) to find the new time taken for a wave to travel the length of the spring. Let \(v_1 = \frac{L}{t_1}\) and \(v_2 = \frac{1.10L}{t_2}\), where \(t_1 = 4.00\,\text{s}\) and \(t_2\) is the time taken to travel the increased length. Since the mass per unit length is constant, the wave speed remains the same, that is, \(v_1 = v_2\).

Solve for the new time taken

Using the wave speed equation, we have: \(\frac{L}{t_1} = \frac{1.10L}{t_2}\) Now we solve for \(t_2\): \(t_2 = t_1 \cdot \frac{1.10L}{L}\) Substitute the given values: \(t_2 = 4.00\,\text{s} \cdot \frac{1.10L}{L}\) Simplify the expression: \(t_2 = 4.4\,\text{s}\) So, after the length of the spring is increased by \(10.0 \%\), the wave takes \(4.4\,\text{s}\) to travel from one end to the other.