Question

A heavy ball of mass $M$ is suspended from the ceiling of a car by a light string of mass $m$ $\left( {m \ll M} \right)$ . When the car is at rest, the speed of transverse waves in the string is $60m{s^{ - 1}}$ . When the car has acceleration $a$ , the wave-speed increases to $60.5m{s^{ - 1}}$ .The value of $a$ , in terms of gravitational acceleration $g$ , is closest to :
(A) $\dfrac{g}{5}$
(B) $\dfrac{g}{{20}}$
(C) $\dfrac{g}{{10}}$
(D) $\dfrac{g}{{30}}$

Answer 1

Hint
To find out the value of $a$ , in terms of acceleration due to gravity $g$ , we have to use the velocity of the wave formula to get the relation when the speed of the transverse wave is $60m{s^{ - 1}}$ . Then again we can use the same formula for the situation when the speed increases to $60.5m{s^{ - 1}}$ . Taking the ratio and simplifying, we get the answer.
Formula Used: In this solution we will be using the following formula,
$v = \sqrt {\dfrac{T}{\mu }}$
Where, T is the tension in the string, v is the velocity of the wave and μ is the linear density of the string.

Complete step by step answer
According to the question, the mass of the heavy ball is $M$ and that of the string is $m$ . It is given in the question that, when the car is at rest, it has a velocity of $60m{s^{ - 1}}$ and when the car is accelerating, the velocity becomes $60.5m{s^{ - 1}}$ .
When we apply the wave velocity formula here, we get,
$v = \sqrt {\dfrac{T}{\mu }}$
The tension in the string will be due to the weight of the mass $M$ . We can neglect the mass of the string since it is given $\left( {m \ll M} \right)$ .
When we substitute the value $v = 60m{s^{ - 1}}$ in the formula, we get
$60 = \sqrt {\dfrac{{Mg}}{\mu }}$ .......(1)
For the second case the car accelerates with an acceleration $a$ , this acceleration is horizontal. The acceleration due to gravity is acting in a downward direction. So $a$ and $g$ are perpendicular to one another.
So when the car is accelerating, the weight of the mass $M$ will be $Mg'$ . Where
$g' = \sqrt {{a^2} + {g^2}}$
So for this case $v = 60.5m{s^{ - 1}}$
Therefore, substituting these in the same formula, we get,
$60.5 = \sqrt {\dfrac{{M\sqrt {{a^2} + {g^2}} }}{\mu }}$ ......(2)
When we divide the equation (2) by equation (1), we get,
$\dfrac{{60.5}}{{60}} = \dfrac{{\sqrt {\dfrac{{M\sqrt {{a^2} + {g^2}} }}{\mu }} }}{{\sqrt {\dfrac{{Mg}}{\mu }} }}$
When we simplify the $\mu$ and $M$ gets cancelled from the numerator and denominator. So we get,
$\dfrac{{60.5}}{{60}} = \dfrac{{\sqrt {\sqrt {{a^2} + {g^2}} } }}{{\sqrt g }}$
Squaring on both sides,
${\left( {\dfrac{{60.5}}{{60}}} \right)^2} = \dfrac{{\sqrt {{a^2} + {g^2}} }}{g}$
When we square both side again,
${\left( {\dfrac{{60.5}}{{60}}} \right)^4} = \dfrac{{{a^2} + {g^2}}}{{{g^2}}}$
Now to find the acceleration we rearrange the equation as,
$\Rightarrow {a^2} = {g^2} \times \dfrac{{{{\left( {60.5} \right)}^4}}}{{{{60}^4}}} - {g^4}$
Now on taking ${g^2}$ common and then taking root on both the sides we get,
$a = g\sqrt {{{\left( {\dfrac{{60.5}}{{60}}} \right)}^4} - 1}$
On calculating this is equal to,
$\Rightarrow a = 0.1837g$ $= \dfrac{g}{{5.44}}$
Which is approximately equal to $\dfrac{g}{5}$ .
Thus, the correct answer is option (A).

Note
When the car is accelerating, the gravity acting on the mass will be the resultant of the acceleration due to gravity and the acceleration of the car. It is given as,
$g' = \sqrt {{a^2} + {g^2} + 2ag\cos \theta }$
Since the angle between them is $90^\circ$ , the third term becomes zero. So the resultant becomes,
$g' = \sqrt {{a^2} + {g^2}}$