A particle starts SHM from the mean position. Its amplitude is a and total energy E . At one instant its kinetic energy is 3(E/4) . Its displacement at that instant is:

Question

A particle starts SHM from the mean position. Its amplitude is a and total energy E . At one instant its kinetic energy is 3(E/4) . Its displacement at that instant is: (A)  (a/√2)  (B)  (a/2)  (C)  (a/√( (3)2 ))  (D)  (a/√3)

Tardigrade Admin · Accepted Answer

k=(1/2)mω 2(a2-y2) (3/4)E=(1/2)mω 2(a2-y2) (3/4)( (1/2)mω 2a2 )=(1/2)mω 2(a2-y2) y2=a2-(3/4)a2 =(a2/4) ⇒ y=(a/2)

Q. A particle starts SHM from the mean position. Its amplitude is a and total energy $E$ . At one instant its kinetic energy is $3 \frac{E}{4}$ . Its displacement at that instant is:

$\frac{a}{2}$

$\frac{a}{2}$

$\frac{a}{( \frac{3}{2} )}$

$\frac{a}{3}$

$a$

$k = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$
$\frac{3}{4} E = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$
$\frac{3}{4} (\frac{1}{2} m ω^{2} a^{2}) = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$
$y^{2} = a^{2} - \frac{3}{4} a^{2}$
$= \frac{a ^{2}}{4}$
$\Rightarrow$ $y = \frac{a}{2}$

Q. A particle starts SHM from the mean position. Its amplitude is a and total energy E . At one instant its kinetic energy is 34E​ . Its displacement at that instant is:

2​a​

2a​

(23​)​a​

3​a​

a

k=21​mω2(a2−y2) 43​E=21​mω2(a2−y2) 43​(21​mω2a2)=21​mω2(a2−y2) y2=a2−43​a2 =4a2​ ⇒ y=2a​

Q. A particle starts SHM from the mean position. Its amplitude is a and total energy $E$ . At one instant its kinetic energy is $3 \frac{E}{4}$ . Its displacement at that instant is:

$\frac{a}{2}$

$\frac{a}{2}$

$\frac{a}{( \frac{3}{2} )}$

$\frac{a}{3}$

$a$

$k = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$
$\frac{3}{4} E = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$
$\frac{3}{4} (\frac{1}{2} m ω^{2} a^{2}) = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$
$y^{2} = a^{2} - \frac{3}{4} a^{2}$
$= \frac{a ^{2}}{4}$
$\Rightarrow$ $y = \frac{a}{2}$