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PhysicsSHMMiscellaneousEasy2 minPYQ_2020

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A mass is suspended from a vertical spring which is executing S.H.M. of frequency $5 Hz$ . The spring is unstretched at the highest point of oscillation. The maximum speed of the mass is [acceleration due to gravity $g = 10 m s^{- 2}$ ]

Options:

Answer:

Solution:

$T = 2 π \sqrt{\frac{m}{k}}$

$n = \frac{1}{2 π} \sqrt{\frac{k}{m}}$

$25 = \frac{1}{4 π^{2}} \frac{k}{m}$

$k = 100 π^{2} m$

$k A = m g$

$A = \frac{m g}{k}$

$V_{max} = ω A$
$= \frac{2 π}{T} A$

$= 2 π n A$

$= \frac{2 π \times 5 \times m g}{k}$
$V_{max} = \frac{10 π \times m \times 10}{100 π^{2} m} = \frac{1}{π} m s^{- 1}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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