Question

A body executes a SHM under the influence of one force with a time period of $3\;s$ and the same body executes a SHM with a time period of $4\;s$ under the influence of a second force. If both the forces are acting simultaneously then, what is the time period of the same body?
A. $2\sqrt 3 \;s$
B. $7\;s$
C. $5\;s$
D. $2.5\;s$

Answer 1

Hint: Given that a body executes simple harmonic motion (SHM) under the influence of some force over a period of $3\;s$, the same body executes SHM under influence of other force over a period of $4\;s$ and when the combination of the both forces acted on the same body, the produced acceleration will get added. By using the formula for acceleration of the body under SHM, the period value can be calculated. In relation with the two forces and the combined force action, the total period can be obtained.

Useful formula:
By Newton’s second law,
$F = ma$
Where, $F$ is the force applied on the body, $m$ is the mass of the body and $a$ is the acceleration of the body.

Angular frequency, $\omega = \dfrac{{2\pi }}{T}$
Where, $T$ is the period of motion.

In simple harmonic motion the acceleration of body, $a = {\omega ^2}x$
Where, $x$ is the displacement of the body.

Given data:
Time period of body in first condition, ${T_1} = 3\;s$
Time period of body in second condition, ${T_2} = 4\;s$

Step by step solution:
In first condition, the force applied on body is ${F_1}$
By Newton’s law, ${F_1} = m{a_1}$
Where, $m$ is the mass of the body and ${a_1}$ is the acceleration of the body in first condition.
The acceleration of the body in SHM, ${a_1} = {\omega _1}^2x$

Angular frequency in first condition, ${\omega _1} = \dfrac{{2\pi }}{{{T_1}}}$
Where, ${T_1}$ is the period of motion in first condition.
Substituting the value of angular frequency in acceleration, we get
${a_1} = {\left( {\dfrac{{2\pi }}{{{T_1}}}} \right)^2} \times x \\\ {a_1} = \dfrac{{{{\left( {2\pi } \right)}^2}}}{{{T_1}^2}} \times x\;...........................................\left( 1 \right) \\\$

In second condition, the force applied on body is ${F_2}$
By Newton’s law, ${F_2} = m{a_2}$
Where, $m$ is the mass of the body and ${a_2}$ is the acceleration of the body in first condition.
The acceleration of the body in SHM, ${a_2} = {\omega _2}^2x$

Angular frequency in second condition, ${\omega _2} = \dfrac{{2\pi }}{{{T_2}}}$
Where, ${T_2}$ is the period of motion in first condition.
Substituting the value of angular frequency in acceleration, we get
${a_2} = {\left( {\dfrac{{2\pi }}{{{T_2}}}} \right)^2} \times x \\\ {a_2} = \dfrac{{{{\left( {2\pi } \right)}^2}}}{{{T_2}^2}} \times x\;...........................................\left( 2 \right) \\\$
Dividing equation (1) by (2),
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{\left( {\dfrac{{{{\left( {2\pi } \right)}^2}}}{{{T_1}^2}} \times x} \right)}}{{\left( {\dfrac{{{{\left( {2\pi } \right)}^2}}}{{{T_2}^2}} \times x} \right)}} \\\ \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{\left( {\dfrac{1}{{{T_1}^2}}} \right)}}{{\left( {\dfrac{1}{{{T_2}^2}}} \right)}} \\\ \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{T_2}^2}}{{{T_1}^2}} \\\$
Substitute the values of ${T_1}$ and ${T_2}$ in the above equation, we get
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{4^2}}}{{{3^2}}} \\\ \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{16}}{9} \\\ {a_2} = \dfrac{{9{a_1}}}{{16}} \\\$

In third condition, the force applied $F = {F_1} + {F_2}$
The mass of the body is same in all the conditions; thus, the acceleration should get added $a = {a_1} + {a_2}$
Substitute the value ${a_2}$ in above equation, we get
$a = {a_1} + \dfrac{{9{a_1}}}{{16}} \\\ a = \dfrac{{16{a_1} + 9{a_1}}}{{16}} \\\ a = \dfrac{{25{a_1}}}{{16}} \\\$
Since, the acceleration of SHM is $a = {\omega _3}^2x$
And angular frequency, ${\omega _3} = \dfrac{{2\pi }}{{{T_{net}}}}$
Where, ${\omega _3}$ is the angular frequency in third condition, $x$ is the displacement of the body and ${T_{net}}$ is the time period in third condition.
Hence,
${\omega _3}^2x = \dfrac{{25{a_1}}}{{16}}$
Substitute the values of ${\omega _3}$ and ${a_1}$ in above equation, we get
${\left( {\dfrac{{2\pi }}{{{T_{net}}}}} \right)^2}x = \dfrac{{25 \times \left( {\dfrac{{{{\left( {2\pi } \right)}^2}}}{{{T_1}^2}} \times x} \right)}}{{16}} \\\ {\left( {\dfrac{{2\pi }}{{{T_{net}}}}} \right)^2} = \dfrac{{25}}{{16}} \times {\left( {\dfrac{{2\pi }}{{{T_1}}}} \right)^2} \\\$
Taking square root on both sides,
$\left( {\dfrac{{2\pi }}{{{T_{net}}}}} \right) = \dfrac{5}{4} \times \left( {\dfrac{{2\pi }}{{{T_1}}}} \right) \\\ {T_{net}} = {T_1} \times \dfrac{4}{5} \\\$
Substitute the value of ${T_1}$ in above equation, we get
${T_{net}} = 3\;s \times \dfrac{4}{5} \\\ {T_{net}} = \dfrac{{12}}{5}\;s \\\ {T_{net}} = 2.4\;s \\\ {T_{net}} \simeq 2.5\;s \\\$
Hence, the option (D) is correct.

Note: In the first two conditions, the force applied induced the simple harmonic motion on the body with the respective time periods. In the third condition, the force applied in the first two conditions will get added up which leads to the result of the sum of acceleration produced in the first two conditions. By applying the angular frequency formula, the resultant time is calculated.