# Energy in Shm A particle on a spring execute Shm.if the total energy of the particle is double then B) the maximum speed of the particle will increase by the factor of ?? a) 4.                  (B) √8              (c) 2.          (D) √2

Avishek Bhagat
3 years ago
Dear Student,
According to the given problem

The time period (T) of vibrations varies inversely as the square root of the force constant (k) of the spring.
2) The time period (T) of vibrations varies directly as the square root of the mass (m) of the body attached to the string
.So the time period of an object of mass m on a spring executes a simple harmonic motion is given by, So, T = 2π √m/k Thus in simple harmonic motion, the time period of the oscillation is independent of the amplitude. If the mass m of the particle is doubled, then the time period (T ') of the simple harmonic motion would be,
T ' = 2π √2m/k
So, T '/ T = (2π √2m/k)/( 2π √m/k)= √2T ' = √2T
From the above observation, we conclude that the period of oscillation will change by a factor of √2

Yuvraj Singh
Rohan Anand
22 Points
2 years ago
We know, in shm all energy is converted into K.E at mean position.
E = 1/2 mv² -(i)
If energy is doubled
2E = 1/2 m(v')² -(ii)
On divinding, i and ii
1/2 = v²/v'²
v'² = 2v²
v' = √2v
Hence, velocity is √2 times.
HOPE IT HELPED!