JEE Main 2017 · 8 Apr · Shift 1 · Physics Q3

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A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is :

\dfrac14

\dfrac1{2\sqrt2}

\dfrac12

2 Hz

Answer and worked solution

Official answer: C

$f=\dfrac1{2\pi}\sqrt{\dfrac km}$ . Two springs in parallel give $2k$ , and the mass is $8m$ :
$f'=f\sqrt{\frac{2}{8}}=\frac12\text{ Hz}.$

Topics:Spring-mass systems

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JEE Main 2017 · 8 Apr · Shift 1, Physics question 3