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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Physics
Chapter
Oscillations

BITSAT 2019 · Physics · Oscillations

Frequency of oscillation of a body is 5 Hz when a force F _ 1 is...

Frequency of oscillation of a body is 55 Hz when a force F1{F}_{1} is applied and 1212 Hz when another force F2{F}_{2} is applied. If both forces F1{F}_{1} and F2{F}_{2} are applied together, then frequency of oscillation of the body will be

Options

  1. A1313 Hz
  2. B169169 Hz
  3. C6262 Hz
  4. D5252 Hz

Correct answer

A. 1313 Hz

Step-by-step solution

According to the question,

F1=−k1x{F}_{1} = - {k}_{1}x and F2=−k2x{F}_{2} = - {k}_{2}x

∴∴ Frequency of oscillation

v1=12πk1m{v}_{1} = \dfrac{1}{2\pi}\sqrt{\dfrac{k1}{m}}

5=12πk1m5 = \dfrac{1}{2\pi}\sqrt{\dfrac{k1}{m}}

⇒k1=100π2m⇒ {k}_{1} = 100{\pi}^{2}m

Similarly, 12=12πk2m12 = \dfrac{1}{2\pi}\sqrt{\dfrac{k2}{m}}

⇒k2=576π2m⇒ {k}_{2} = 576{\pi}^{2}m

Now, F=F1+F2=−(k1+k2)xF = {F}_{1} + {F}_{2} = - \left({k}_{1} + {k}_{2}\right)x

∴∴ Frequency of oscillation,

v=12πk1+k2m=12π100π2m+576π2mmv = \dfrac{1}{2\pi}\sqrt{\dfrac{{k}_{1} + {k}_{2}}{m}} = \dfrac{1}{2\pi}\sqrt{\dfrac{100{\pi}^{2}m + 576{\pi}^{2}m}{m}}

⇒v=13Hz⇒ v = 13Hz

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