A heavy ball hangs from the ceiling of a motor car by a light...

Let $m$ be the mass of the heavy ball and $\mu$ be the mass per unit length of the string.

When the car is stationary, the tension in the string is $T_1 = mg$.

The speed of the transverse pulse is given by $v_1 = \sqrt{\dfrac{T_1}{\mu}} = \sqrt{\dfrac{mg}{\mu}} = 60 \text{ cm s}^{-1}$.

When the car accelerates horizontally with acceleration $a$, the effective acceleration due to gravity becomes $g_{\text{eff}} = \sqrt{g^2 + a^2}$.

The new tension in the string is $T_2 = m\sqrt{g^2 + a^2}$.

The new speed of the transverse pulse is $v_2 = \sqrt{\dfrac{T_2}{\mu}} = \sqrt{\dfrac{m\sqrt{g^2 + a^2}}{\mu}} = 62 \text{ cm s}^{-1}$.

Taking the ratio of the speeds:

$\dfrac{v_2}{v_1} = \sqrt{\dfrac{T_2}{T_1}} = \sqrt{\dfrac{\sqrt{g^2 + a^2}}{g}} = \left(1 + \dfrac{a^2}{g^2}\right)^{1/4}$

Substituting the given values:

\left(\dfrac{62}{60}\right)^4 = 1 + \dfrac{a^2}{g^2}$<br><br>\left(\dfrac{31}{30}\right)^4 = 1 + \dfrac{a^2}{10^2}$

$1.140 \approx 1 + \dfrac{a^2}{100}$

$\dfrac{a^2}{100} \approx 0.140$

$a^2 \approx 14.0$

$a = \sqrt{14.0} \approx 3.74 \text{ m s}^{-2}$