Kinetic energy is KE\(=\frac { 1 }{ 2 } { mv }_{ x }^{ 2 }\)
Multiply numerator and denominator by m
\(KE=\frac { 1 }{ 2m } { m^{ 2 }v }_{ x }^{ 2 }=\frac { 1 }{ 2m } \left( { mv }_{ x } \right) ^{ 2 }=\frac { 1 }{ 2m } { P }_{ x }^{ 2 }\)
where, P_x is the linear momentum of the particle executing simple harmonic motion.
Total energy can be written as sum of kinetic energy and potential energy, therefore, from equation (10.73) and also from equation (10.75), we get
E = KE + U(x) \(=\frac { 1 }{ 2m } { P }_{ x }^{ 2 }+\frac { 1 }{ 2m } { m\omega ^{ 2 }v }_{ x }^{ 2 }\) = constant

Since, \(T=\propto \sqrt { l } \)
Therefore,
T = constant\(\sqrt{l}\)
\(\frac { { T }_{ f } }{ { T }_{ i } } =\sqrt { \frac { l+\frac { 44 }{ 100 } l }{ l } } =\sqrt { 1.44 } =1.2\)
Therefore, T_f = 1.2 T_i = T_i + 20% T_i

k₁ = 1 N m−¹, k₂ = 2 N m⁻¹
k_p = k₁ + k₂ N m⁻¹
k_p = 1 + 2 = 3 N m⁻¹
k_p > k₁ and k_p > k₂
Therefore, the effective spring constant is greater than both k₁ and k₂.

Let the number of heart beats measured be f. Since the time period is inversely proportional to the heart beat, then
\(f=\frac { 1 }{ T } =\frac { 1 }{ 0.8 } ={ 1.25 }s^{ -1 }\)
One minute is 60 second,
(1 second = \(\frac{1}{60}\) minute \(\Rightarrow\) 1 s⁻¹ = 60 min⁻¹)
f =1.25 s⁻¹ \(\Rightarrow\) f = 1.25\(\times\)60 min⁻¹ = 75 beats per minute

Let us first calculate the stiffness constant of the spring balance by using equation (10.29),
\(K=\frac { mg }{ l } =\frac { 25\times 11.5 }{ 0.25 } ={ 1150 }Nm^{ -1 }\)
The time period of oscillations is given by
\(T=2\pi \sqrt { \frac { M }{ k } } \)where M is the mass of the body.
Since, M is unknown, rearranging, we get 
\(M=\frac { k{ T }^{ 2 } }{ { 4\pi }^{ 2 } } =\frac { \left( 1150 \right) { \left( 0.5 \right) }^{ 2 } }{ { 4\pi }^{ 2 } } =7.3kg\)
The gravitational force acting on the body is W = Mg = 7.3\(\times\)11.5 = 83.95 N ≈ 84 N