A body of mass 5 gm is executing S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm/sec. Its velocity will be 50 cm/sec at a distance

[CPMT 1976]

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Solution

The correct option is C

It is given $v_{m a x}$ = 100 m /sec, a = 10 cm.

$\Rightarrow v_{m a x} = a ω \Rightarrow ω = \frac{100}{10} = 10 r a d / s e c$

Hence v = $ω \sqrt{a^{2} - y^{2}} \Rightarrow 50 = 10 \sqrt{(10)^{2} - y^{2}}$

$\Rightarrow$ y = $5 \sqrt{3}$ cm

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A body of mass 5 gm is executing S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm/sec. Its velocity will be 50 cm/sec at a distance [CPMT 1976]

The correct option is C It is given vmax = 100 m /sec, a = 10 cm. ⇒vmax=aω⇒ω=10010=10rad/sec Hence v = ω√a2−y2⇒50=10√(10)2−y2 ⇒ y = 5√3 cm

A body of mass 5 gm is executing S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm/sec. Its velocity will be 50 cm/sec at a distance

[CPMT 1976]

The correct option is C

It is given $v_{m a x}$ = 100 m /sec, a = 10 cm.

$\Rightarrow v_{m a x} = a ω \Rightarrow ω = \frac{100}{10} = 10 r a d / s e c$

Hence v = $ω \sqrt{a^{2} - y^{2}} \Rightarrow 50 = 10 \sqrt{(10)^{2} - y^{2}}$

$\Rightarrow$ y = $5 \sqrt{3}$ cm