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Step-by-Step Solution
Step 1: Identify the formula for wave speed in a string
For a string under tension $T$ and having linear mass density $\mu$, the speed of transverse waves $v$ is given by:
$v = \sqrt{\frac{T}{\mu}}$
Step 2: Express wave speed when the car is at rest
When the car is at rest, the only force acting on the suspended mass $M$ (vertically) is its weight $Mg$. Thus, the tension in the string, $T_{\text{rest}}$, is:
$T_{\text{rest}} = Mg$
The corresponding wave speed is given as 60 m/s, so:
$60 = \sqrt{\frac{Mg}{\mu}}$
Step 3: Express wave speed when the car accelerates
When the car accelerates horizontally with acceleration $a$, the tension in the string becomes the resultant of two perpendicular forces: the weight $Mg$ (downward) and the inertial force $Ma$ (horizontal). Hence:
$T_{\text{acc}} = M \sqrt{g^2 + a^2}
The new wave speed $v_{\text{acc}}$ is given as 60.5 m/s, thus:
$60.5 = \sqrt{\frac{M \sqrt{g^2 + a^2}}{\mu}}
Step 4: Form a ratio to eliminate common factors
To find the relation between $g$ and $a$, take the ratio of the two speeds and square it to relate the tensions:
$\frac{v_{\text{acc}}}{v_{\text{rest}}} = \sqrt{ \frac{T_{\text{acc}}}{T_{\text{rest}}} }
\quad\Longrightarrow\quad
\left(\frac{60.5}{60}\right)^2 = \frac{T_{\text{acc}}}{T_{\text{rest}}}
Since $T_{\text{rest}} = Mg$ and $T_{\text{acc}} = M \sqrt{g^2 + a^2}$, we have:
$\left(\frac{60.5}{60}\right)^2 = \frac{M \sqrt{g^2 + a^2}}{Mg} = \frac{\sqrt{g^2 + a^2}}{g}
Step 5: Solve for the acceleration $a$
$\frac{\sqrt{g^2 + a^2}}{g} = \left(\frac{60.5}{60}\right)^2
Calculate the numerical ratio on the right side:
$\left(\frac{60.5}{60}\right)^2 \approx \frac{3660.25}{3600} \approx 1.0167
So, $\sqrt{g^2 + a^2} = 1.0167\,g \quad\Longrightarrow\quad g^2 + a^2 = (1.0167)^2\,g^2
$(1.0167)^2 \approx 1.0336 \quad\Longrightarrow\quad g^2 + a^2 = 1.0336\,g^2
a^2 = (1.0336 - 1)\,g^2 = 0.0336\,g^2
\quad\Longrightarrow\quad
a = g\,\sqrt{0.0336}\approx 0.183\,g
This is close to $0.2\,g$, which corresponds to $\frac{g}{5}$.
Step 6: Conclude the correct answer
The value of the car's acceleration $a$ that makes the wave speed increase from 60 m/s to 60.5 m/s is approximately $\frac{g}{5}$.
