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

k₁ and k₂ attached to a mass m as shown in figure. The results can be generalized to any number of springs in parallel.

Let the force F be applied towards right as shown in figure. In this case, both the springs elongate or compress by the same amount of displacement. Therefore, net force for the displacement of mass m is
F = - k_px  ..(1)
where kp is called effective spring constant.
Let the first spring be elongated by a displacement x due to force F₁ and second spring be elongated by the same displacement x due to force F₂' then the net force
F = - k₁x - k₂x ...(2)
Equating equations (2) and (1), we get
k_p= k₁ + k_{2     ...}(3)
Generalizing, for n springs connected in parallel
\({ k }_{ P }=\sum _{ i=1 }^{ n }{ { k }_{ i } } \) ....(4)
If all spring constants are identical i.e.,k₁ = k₂ = ... = kn = k then
k_p = n k    ...(5)
This implies that the effective spring constant increases by a factor n. Hence, for the springs in parallel connection, the effective spring constant is greater than individual spring constant.

T=2\(\pi \sqrt { \frac { l }{ g } } \)
(i) As the water flows out of the sphere, the times period first increases and then decreases.
(ii) Initially when the sphere is completely filled with water, its C.G lies at its center.
(iii) As water flows out, the C.G begins to shift below the centre of the sphere the effective length of the pendulum increases & hence its time period increase when the sphere becomes more than half empty, its C.G begins to rise up the effective length of the pendulum increases and true period T decreases.
(iv) When the entire water is drained out of the sphere, the C.G is once again shifted to centre of the sphere and the time period T attains its initial value.

V_max= \({ \omega }_{ A }=\frac { 2\pi }{ T } A\)
T=\(\frac { 2\pi A }{ { V }_{ max } } =\frac { 2\times 22\times 7\times { 10 }^{ -3 } }{ 7\times 4.4 } \)
= 0.01s

\(\frac{potential \ energy}{Total \ energy}=\frac { \frac { 1 }{ 2 } m{ \omega }^{ 2 }{ y }^{ 2 } }{ \frac { 1 }{ 2 } m{ \omega }^{ 2 }{ a }^{ 2 } } \)
\(\frac { { y }^{ 2 } }{ { a }^{ 2 } } =\frac { \left( \frac { a }{ 2 } \right) }{ a } =\frac { 1 }{ 4 } \)
=1:4

Time period T remains same, this is cos the period of oscillation of a liquid in a U-tube does not depend on the density of the liquid.

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