Oscillations (SHM) · JEE Main previous year questions

2019 9 questions

1. Q522019 · 12 Apr 2019 · Shift 2

   A spring whose unstretched length is l has a force constant k. The spring is cut into two pieces of unstretched lengths $l_1$ and $l_2$ where, $l_1=n{l}_2$ and n is an integer. The ratio $k_1/k_2$ of the corresponding force constant, $k_1$ and $k_2$ will be :

   1. A.$\frac{1}{n^2}$
   2. B.$\frac{1}{n}$
   3. C.$n^2$
   4. D.n
   Show answer

   Answer: B

2. Q562019 · 10 Apr 2019 · Shift 1

   The displacement of a damped harmonic oscillator is given by $x(t)=e^{-0.1t}\cos (10\pi t+f)$. Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to :

   1. A.27 s
   2. B.13 s
   3. C.7 s
   4. D.4 s
   Show answer

   Answer: C

3. Q362019 · 9 Apr 2019 · Shift 1

   A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is 1/16 th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is :

   1. A.$2T\sqrt{\frac{1}{10}}$
   2. B.$4T\sqrt{\frac{1}{14}}$
   3. C.$4T\sqrt{\frac{1}{15}}$
   4. D.$2T\sqrt{\frac{1}{14}}$
   Show answer

   Answer: C

4. Q602019 · 8 Apr 2019 · Shift 2

   A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1/1000 of the original amplitude is close to :-

   1. A.100 s
   2. B.10 s
   3. C.20 s
   4. D.50 s
   Show answer

   Answer: C

5. Q432019 · 12 Jan 2019 · Shift 1

   A simple pendulum, made of a string of length $\ell$ and a bob of mass m, is released from a small angle ${\theta }_0$. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle ${\theta }_1$. Then M is given by :

   1. A.$\frac{m}{2}\left(\frac{{\theta }_0+{\theta }_1}{{\theta }_0-{\theta }_1}\right)$
   2. B.$\frac{m}{2}\left(\frac{{\theta }_0-{\theta }_1}{{\theta }_0+{\theta }_1}\right)$
   3. C.$m\left(\frac{{\theta }_0+{\theta }_1}{{\theta }_0-{\theta }_1}\right)$
   4. D.$m\left(\frac{{\theta }_0-{\theta }_1}{{\theta }_0+{\theta }_1}\right)$
   Show answer

   Answer: C

6. Q362019 · 12 Jan 2019 · Shift 2

   A simple harmonic motion is represented by : $y=5(\sin 3\pi t+\sqrt{3}\cos 3\pi t)\mathrm{cm}$ The amplitude and time period of the motion are :

   1. A.$10\mathrm{cm},\frac{3}{2}\mathrm{s}$
   2. B.$5\mathrm{cm},\frac{2}{3}\mathrm{s}$
   3. C.$5\mathrm{cm},\frac{3}{2}\mathrm{s}$
   4. D.$10\mathrm{cm},\frac{2}{3}\mathrm{s}$
   Show answer

   Answer: D

7. Q382019 · 11 Jan 2019 · Shift 2

   A pendulum is executing simple harmonic motion and its maximum kinetic energy is $K_1$. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is $K_2$. Then :

   1. A.$K_2=\frac{K_1}{2}$
   2. B.$K_2=2K_1$
   3. C.$K_2=K_1$
   4. D.$K_2=\frac{K_1}{4}$
   Show answer

   Answer: B

8. Q392019 · 11 Jan 2019 · Shift 2

   The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of simple pendulum on the Earth is $2\text{ s}$. The period of oscillation of the same pendulum on the planet would be :

   1. A.$\frac{\sqrt{3}}{2}\text{ s}$
   2. B.$\frac{3}{2}\text{ s}$
   3. C.$\frac{2}{\sqrt{3}}\text{ s}$
   4. D.$2\sqrt{3}\text{ s}$
   Show answer

   Answer: D

9. Q592019 · 11 Jan 2019 · Shift 2

   A simple pendulum of length $1\text{ m}$ is oscillating with an angular frequency $10\text{ rad/s}$. The support of the pendulum starts oscillating up and down with a small angular frequency of $1\text{ rad/s}$ and an amplitude of $10^{-2}\text{ m}$. The relative change in the angular frequency of the pendulum is best given by :

   1. A.$1\text{ rad/s}$
   2. B.$10^{-3}\text{ rad/s}$
   3. C.$10^{-1}\text{ rad/s}$
   4. D.$10^{-5}\text{ rad/s}$
   Show answer

   Answer: B