Differential Equation of S.H.M.

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   i. In a linear S.H.M., the force is directed towards the mean position and its magnitude is directly proportional to the displacement of the body from mean position. $\therefore f\propto -x$ $\therefore f=-kx\qquad....(1)$ where, k is force constant and x is displacement from the mean position. ii. According to Newton's second law of motion, $f=ma\qquad....(2)$ From equations (1) and (2), $ma=-kx\qquad....(3)$ iii. The velocity of the particle is given by, $v=\dfrac{dx}{dt}$, $\therefore$ Acceleration, $a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}\qquad....(4)$ iv. Substituting equation (4) in equation (3), $m\dfrac{d^2x}{dt^2}=-kx$ $\therefore \dfrac{d^2x}{dt^2}+\dfrac{k}{m}x=0$ v. Substituting $\dfrac{k}{m}=\omega^2$, where ω is the angular frequency, $\therefore \dfrac{d^2x}{dt^2}+\omega^2x=0$ This is the differential equation of linear S.H.M.

2. 2.↑ back to the question

   Definition: Linear S.H.M. is defined as the linear periodic motion of a body, in which force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position. Differential equation of linear S.H.M.: [From 5.4, Q1]

   i. In a linear S.H.M., the force is directed towards the mean position and its magnitude is directly proportional to the displacement of the body from mean position. $\therefore f\propto -x$ $\therefore f=-kx\qquad....(1)$ where, k is force constant and x is displacement from the mean position. ii. According to Newton's second law of motion, $f=ma\qquad....(2)$ From equations (1) and (2), $ma=-kx\qquad....(3)$ iii. The velocity of the particle is given by, $v=\dfrac{dx}{dt}$, $\therefore$ Acceleration, $a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}\qquad....(4)$ iv. Substituting equation (4) in equation (3), $m\dfrac{d^2x}{dt^2}=-kx$ $\therefore \dfrac{d^2x}{dt^2}+\dfrac{k}{m}x=0$ v. Substituting $\dfrac{k}{m}=\omega^2$, where ω is the angular frequency, $\therefore \dfrac{d^2x}{dt^2}+\omega^2x=0$ This is the differential equation of linear S.H.M.

3. 3.↑ back to the question

   [From 5.4, Q1 (Definition only)]

   i. In a linear S.H.M., the force is directed towards the mean position and its magnitude is directly proportional to the displacement of the body from mean position. $\therefore f\propto -x$ $\therefore f=-kx\qquad....(1)$ where, k is force constant and x is displacement from the mean position. ii. According to Newton's second law of motion, $f=ma\qquad....(2)$ From equations (1) and (2), $ma=-kx\qquad....(3)$ iii. The velocity of the particle is given by, $v=\dfrac{dx}{dt}$, $\therefore$ Acceleration, $a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}\qquad....(4)$ iv. Substituting equation (4) in equation (3), $m\dfrac{d^2x}{dt^2}=-kx$ $\therefore \dfrac{d^2x}{dt^2}+\dfrac{k}{m}x=0$ v. Substituting $\dfrac{k}{m}=\omega^2$, where ω is the angular frequency, $\therefore \dfrac{d^2x}{dt^2}+\omega^2x=0$ This is the differential equation of linear S.H.M.

4. 4.↑ back to the question

   Given: $v_{max}=0.16$ m/s, $a_{max}=0.64$ m/s$^2$ To find: Period (T) Formulae: i. $v_{max}=A\omega$ ii. $a_{max}=A\omega^2$ iii. $\omega=\dfrac{2\pi}{T}$ Calculation: From formulae, $\dfrac{a_{max}}{v_{max}}=\omega=\dfrac{2\pi}{T}$ $\therefore T=\dfrac{2\pi\times v_{max}}{a_{max}}=\dfrac{2\times3.142\times0.16}{0.64}$ $=1.571$ s Ans: Period of particle is 1.571 s.

5. 5.↑ back to the question

   Given: T = 2 s, A = 10 cm To find: Acceleration (a) Formula: $a=\omega^2x=\left(\dfrac{2\pi}{T}\right)^2x$ (in magnitude) Calculation: Particle is 4 cm from its positive extreme position, hence x = 10 − 4 = 6 cm From formula, $a=\left(\dfrac{2\pi}{2}\right)^2\times6$ $=6\pi^2=6\times(3.142)^2$ = antilog {log(6) + 2 log(3.142)} = antilog {0.7782 + 2(0.4972)} = antilog {0.7782 + 0.9944} = antilog {1.7726} = 59.23 cm/s$^2$ Ans: The acceleration of the particle when at 4 cm from its extreme is 59.23 cm/s$^2$.

6. 6.↑ back to the question

   Given: a = 1 cm, $t=\dfrac{\pi}{3}$ s, $\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{2\pi}=1$ rad/s To find: Displacement (x) Formula: $x=A\cos\omega t$ Calculation: From formula, $x=1\cos\left(1\times\dfrac{\pi}{3}\right)=\cos\left(\dfrac{\pi}{3}\right)$ $\therefore x=0.5$ cm Ans: The displacement of the particle after $\dfrac{\pi}{3}$ seconds is 0.5 cm.

7. 7.↑ back to the question

   Given: 2A = 20 cm $\therefore$ A = 10 cm, T = 6.28 s, x = 6 cm To find: Velocity (v) Formula: $v=\omega\sqrt{A^2-x^2}$ Calculation: Since, $\omega=\dfrac{2\pi}{T}=\dfrac{2\times3.14}{6.28}=1$ rad/s $\therefore v=(1)\sqrt{(10)^2-(6)^2}=\sqrt{100-36}=8$ cm/s Ans: The velocity of the particle at x = 6 cm is 8 cm/s.

8. 8.↑ back to the question

   Given: $v_{max}=25$ cm/s, $a_{max}=100$ cm/s$^2$ To find: Amplitude (A), Period (T) Formulae: i. $v_{max}=A\omega$ ii. $a_{max}=A\omega^2$ iii. $\omega=\dfrac{2\pi}{T}$ Calculation: From formulae (i) and (ii), $\omega=\dfrac{a_{max}}{v_{max}}=\dfrac{100}{25}=4$ rad/s From formula (iii), $\therefore T=\dfrac{2\pi}{\omega}=\dfrac{2\times3.142}{4}=1.571$ s From formula (i), $\therefore A=\dfrac{v_{max}}{\omega}=\dfrac{25}{4}=6.25$ cm Ans: Amplitude is 6.25 cm and time period is 1.571 s

9. 9.↑ back to the question

   Given: m = 1 kg, k = 16 N/m To find: Angular frequency (ω), frequency of vibration (n) Formulae: i. $\omega=\sqrt{\dfrac{k}{m}}$ ii. $\omega=2\pi n$ Calculation: From formula (i), We have, for S.H.M. $\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{16}{1}}$ $\therefore \omega=4$ rad/s From formula (ii), $n=\dfrac{\omega}{2\pi}=\dfrac{4}{2\pi}=\dfrac{2}{\pi}$ Hz $\therefore n=\dfrac{2}{3.142}=0.6365$ Hz Ans: The angular frequency of the body is 4 rad/s and the frequency of vibration is 0.6365 Hz.

10. 10.↑ back to the question

    Given: $v_{max}=25$ cm/s, $a_{max}=100$ cm/s$^2$ To find: Period (T) Formulae: i. $v_{max}=A\omega$ ii. $a_{max}=A\omega^2$ iii. $\omega=\dfrac{2\pi}{T}$ Calculation: From formula (i), (ii) and (iii), $\dfrac{a_{max}}{v_{max}}=\omega=\dfrac{2\pi}{T}$ $\therefore T=\dfrac{2\pi\times v_{max}}{a_{max}}=\dfrac{2\times3.142\times25}{100}=1.571$ s Ans: Time period of oscillation is 1.571 s.

11. 11.↑ back to the question

    $v_{max}=\omega A$ and $a_{max}=\omega^2A$ $\therefore \dfrac{a_{max}}{v_{max}}=\dfrac{\omega^2A}{\omega A}=\omega$ $\therefore \dfrac{0.32}{0.08}=\dfrac{2\pi}{T}$ $\therefore T=\dfrac{\pi}{2}=\dfrac{3.142}{2}=1.571$ s