Home
Class 11
PHYSICS
The resultant amplitude due to superposi...

The resultant amplitude due to superposition of three simple harmonic motions `x_(1) = 3sin omega t`,
`x_(2) = 5sin (omega t + 37^(@))` and `x_(3) = - 15cos omega t` is

A

`18`

B

`10`

C

`12`

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To find the resultant amplitude due to the superposition of the three simple harmonic motions given by: 1. \( x_1 = 3 \sin(\omega t) \) 2. \( x_2 = 5 \sin(\omega t + 37^\circ) \) 3. \( x_3 = -15 \cos(\omega t) \) we will follow these steps: ### Step 1: Convert \( x_3 \) to sine form The equation \( x_3 = -15 \cos(\omega t) \) can be converted to sine form using the identity \( \cos(\theta) = \sin(\theta + \frac{\pi}{2}) \): \[ x_3 = -15 \cos(\omega t) = 15 \sin\left(\omega t - \frac{\pi}{2}\right) \] ### Step 2: Rewrite the equations Now we have: - \( x_1 = 3 \sin(\omega t) \) - \( x_2 = 5 \sin(\omega t + 37^\circ) \) - \( x_3 = 15 \sin\left(\omega t - \frac{\pi}{2}\right) \) ### Step 3: Break down \( x_2 \) into components Using the angle addition formula, we can express \( x_2 \): \[ x_2 = 5 \left( \sin(\omega t) \cos(37^\circ) + \cos(\omega t) \sin(37^\circ) \right) \] Calculating \( \cos(37^\circ) \) and \( \sin(37^\circ) \): - \( \cos(37^\circ) \approx \frac{4}{5} \) - \( \sin(37^\circ) \approx \frac{3}{5} \) Thus, we can write: \[ x_2 = 5 \left( \sin(\omega t) \cdot \frac{4}{5} + \cos(\omega t) \cdot \frac{3}{5} \right) = 4 \sin(\omega t) + 3 \cos(\omega t) \] ### Step 4: Combine all components Now we can express \( x_1 \), \( x_2 \), and \( x_3 \) in terms of sine and cosine: \[ x_1 + x_2 + x_3 = (3 + 4) \sin(\omega t) + (3 - 15) \cos(\omega t) \] \[ = 7 \sin(\omega t) - 12 \cos(\omega t) \] ### Step 5: Find the resultant amplitude To find the resultant amplitude \( A \), we can use the formula: \[ A = \sqrt{X^2 + Y^2} \] where \( X = 7 \) (coefficient of \( \sin(\omega t) \)) and \( Y = -12 \) (coefficient of \( \cos(\omega t) \)): \[ A = \sqrt{7^2 + (-12)^2} = \sqrt{49 + 144} = \sqrt{193} \approx 13.89 \] ### Final Result The resultant amplitude is approximately \( 13.89 \, \text{cm} \). ---

To find the resultant amplitude due to the superposition of the three simple harmonic motions given by: 1. \( x_1 = 3 \sin(\omega t) \) 2. \( x_2 = 5 \sin(\omega t + 37^\circ) \) 3. \( x_3 = -15 \cos(\omega t) \) we will follow these steps: ...
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    DC PANDEY ENGLISH|Exercise Level 1 Subjective|39 Videos
  • SIMPLE HARMONIC MOTION

    DC PANDEY ENGLISH|Exercise Level 2 Single Correct|28 Videos
  • SIMPLE HARMONIC MOTION

    DC PANDEY ENGLISH|Exercise Level 1 Assertion And Reason|10 Videos
  • SEMICONDUCTORS AND ELECTRONIC DEVICES

    DC PANDEY ENGLISH|Exercise More than One Option is Correct|3 Videos
  • SOLVD PAPERS 2017 NEET, AIIMS & JIPMER

    DC PANDEY ENGLISH|Exercise Solved paper 2018(JIPMER)|38 Videos
DC PANDEY ENGLISH-SIMPLE HARMONIC MOTION-Level 1 Single Correct
  1. Two simple harmonic motions are given by y(1) = a sin [((pi)/(2))t + p...

    Text Solution

    |

  2. A particle starts performing simple harmonic motion. Its amplitude is ...

    Text Solution

    |

  3. Which of the following is not simple harmonic function ?

    Text Solution

    |

  4. The displacement of a particle varies according to the relation x=4 (c...

    Text Solution

    |

  5. Two pendulums X and Y of time periods 4 s and 4.2s are made to vibrate...

    Text Solution

    |

  6. A mass M is suspended from a massless spring. An additional mass m str...

    Text Solution

    |

  7. Two bodies P and Q of equal masses are suspended from two separate mas...

    Text Solution

    |

  8. A disc of radius R is pivoted at its rim. The period for small oscilla...

    Text Solution

    |

  9. Identify the correct variation of potential energy U as a function of ...

    Text Solution

    |

  10. If the length of a simple pendulum is equal to the radius of the earth...

    Text Solution

    |

  11. The displacement - time (x - t) graph of a particle executing simple h...

    Text Solution

    |

  12. In the figure shown the time period and the amplitude respectively, wh...

    Text Solution

    |

  13. The equation of motion of a particle of mass 1g is (d^(2)x)/(dt^(2)) +...

    Text Solution

    |

  14. The spring as shown in figure is kept in a stretched position with ext...

    Text Solution

    |

  15. The mass and diameter of a planet are twice those of earth. What will ...

    Text Solution

    |

  16. The resultant amplitude due to superposition of three simple harmonic ...

    Text Solution

    |

  17. Two SHMs s(1) = a sin omega t and s(2) = b sin omega t are superimpose...

    Text Solution

    |

  18. The amplitude of a particle executing SHM about O is 10 cm. Then

    Text Solution

    |

  19. A particle is attached to a vertical spring and is pulled down a dista...

    Text Solution

    |

  20. A block of mass 1kg is kept on smooth floor of a truck. One end of a s...

    Text Solution

    |