The amplitude of a particle executing SHM is 4 cm. At the mean position, the speed of the particle is 16 cm / s. The distance of the particle from the mean position at which the speed of the particle becomes 8 √3 cm / sec will be

Question

The amplitude of a particle executing SHM is 4 cm. At the mean position, the speed of the particle is 16 cm / s. The distance of the particle from the mean position at which the speed of the particle becomes 8 √3 cm / sec will be (A) 2 √3 (B) √3 (C) 1 (D) 2

Tardigrade Admin · Accepted Answer

ω=(V max /a) ⇒ (16/4)=4 V=ω √a2-y2 ⇒ 8 √3=4 √16-y2 ⇒ 2 √3=√16-y2 ⇒ 12=16- y 2 ⇒ y 2=4 (squaring both side) y=2 cm

Q. The amplitude of a particle executing $S H M$ is $4 c m$ . At the mean position, the speed of the particle is $16 c m / s$ . The distance of the particle from the mean position at which the speed of the particle becomes $83 c m / sec$ will be

$23$

$3$

1

2

$ω = \frac{V _{m a x}}{a} \Rightarrow \frac{16}{4} = 4$
$V = ω a^{2} - y^{2}$
$\Rightarrow 83 = 4 16 - y^{2}$
$\Rightarrow 23 = 16 - y^{2}$
$\Rightarrow 12 = 16 - y^{2}$
$\Rightarrow y^{2} = 4$ (squaring both side)
$y = 2 c m$

Q. The amplitude of a particle executing SHM is 4cm. At the mean position, the speed of the particle is 16cm/s. The distance of the particle from the mean position at which the speed of the particle becomes 83​cm/sec will be

23​

3​

1

2

ω=aVmax​​⇒416​=4 V=ωa2−y2​ ⇒83​=416−y2​ ⇒23​=16−y2​ ⇒12=16−y2 ⇒y2=4 (squaring both side) y=2cm

Q. The amplitude of a particle executing $S H M$ is $4 c m$ . At the mean position, the speed of the particle is $16 c m / s$ . The distance of the particle from the mean position at which the speed of the particle becomes $83 c m / sec$ will be

$23$

$3$

$ω = \frac{V _{m a x}}{a} \Rightarrow \frac{16}{4} = 4$
$V = ω a^{2} - y^{2}$
$\Rightarrow 83 = 4 16 - y^{2}$
$\Rightarrow 23 = 16 - y^{2}$
$\Rightarrow 12 = 16 - y^{2}$
$\Rightarrow y^{2} = 4$ (squaring both side)
$y = 2 c m$