- #1
vissh
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Time period of SHM !
ellloO ^.^
<Q>A particle executes SHM under the restoring force provided by a spring. The time period is T . If the spring is divided in 2 equal parts and one part is used to continue the SHM, The time period will
<A> remain T <b> become T/2 <c> become 2T <d> become T/(21/2)
>If on a particle of mass "m" at a displacement "x" from the mean position , a force " -kx " acts ;then the particle will be in SHM abt that mean position. [ k = constant]
T = [2 [tex]\pi[/tex] (m)1/2]/(k)1/2
>>Earlier, the particle was acted upon by force of magnitude "kx" when it was "x" away from mean position and thus time period
T =[2 [tex]\pi[/tex](m)1/2]/(k)/2
>>If we use the half part only , still the particle will be acted upon by force of magnitude "kx" and time period T will not change.
>>But the answer is <d> become T/(21/2)
Little help ^.^
{ lol don't know why the Pie symbol is flying above all other words }
ellloO ^.^
Homework Statement
<Q>A particle executes SHM under the restoring force provided by a spring. The time period is T . If the spring is divided in 2 equal parts and one part is used to continue the SHM, The time period will
<A> remain T <b> become T/2 <c> become 2T <d> become T/(21/2)
Homework Equations
>If on a particle of mass "m" at a displacement "x" from the mean position , a force " -kx " acts ;then the particle will be in SHM abt that mean position. [ k = constant]
T = [2 [tex]\pi[/tex] (m)1/2]/(k)1/2
The Attempt at a Solution
>>Earlier, the particle was acted upon by force of magnitude "kx" when it was "x" away from mean position and thus time period
T =[2 [tex]\pi[/tex](m)1/2]/(k)/2
>>If we use the half part only , still the particle will be acted upon by force of magnitude "kx" and time period T will not change.
>>But the answer is <d> become T/(21/2)
Little help ^.^
{ lol don't know why the Pie symbol is flying above all other words }