• Physics 9702 Doubts | Help Page 90

Question 462: [Waves > Stationary waves]
Speed of a transverse wave on a stretched string can be changed by adjusting tension of the string. A stationary wave pattern is set up on stretched string using an oscillator set at frequency of 650 Hz.

How must wave be changed to maintain the same stationary wave pattern if the applied frequency is increased to 750 Hz?
A Decrease the speed of the wave on the string.
B Decrease the wavelength of the wave on the string.
C Increase the speed of the wave on the string.
D Increase the wavelength of the wave on the string.

Reference: Past Exam Paper – June 2013 Paper 11 Q28



Solution 462:

Answer: C.

One complete wave is obtained (as shown in the diagram). This corresponds to 1 wavelength.

The length of the string is unchanged. To maintain the same wave pattern, that is, for only one complete wave to be obtained again, the wavelength of the wave must remain constant.

Speed of wave, v = fλ

The wavelength λ needs to be kept constant and the frequency is increased to 750Hz.

Wavelength λ = v / f

For λ to have the same value, the speed of the wave, v, needs to be increased.

Question 463: [Forces > Hooke’s law]
Spring of unextended length 40 mm is suspended from a fixed point. Load of 16 N is applied to the free end of the spring. This causes spring to extend so that its final length is five times its original length. The spring obeys Hooke’s Law.
What is the energy stored in the spring due to this extension?
A 1.3 J                         B 1.6 J                         C 2.6 J                         D 3.2 J

Reference: Past Exam Paper – June 2013 Paper 13 Q15



Solution 463:

Answer: A.

Original (unextended) length, L = 40mm

Load applied, F = 16N

Extension, e = 5(40) – 40 = 160mm

Energy stored in spring = ½ Fx = 0.5 (16) (160x10^-3) = 1.28 = 1.3J

Question 464: [Kinematics]
Two balls roll from rest down the same slope, but ball X reaches the bottom of the slope in half the time it takes Y. If ball X was accelerating at a ms^-2, then Y's acceleration, in ms^-2, was
A ¼ a                          B ½ a                           C 2 a                            D 4a



Solution 464:
Answer: A.
Consider ball X.
Initial speed, u = 0. Acceleration = a. Time taken = t. Let the distance travelled = s
Equation for uniformly accelerated motion: s = ut + ½ at²
For ball X: s = 0 + 0.5at² = 0.5at²

Now, consider ball Y.
The distance travelled is the same (and equal to s). Initial speed, u = 0. Time taken = 2t. Let the acceleration = x.
Equation for uniformly accelerated motion: s = ut + ½ at²
For ball Y: s = 0 + 0.5x (2t)² = 2xt²

Since the distance s is the same for both cases, we can equate both equations.
2xt² = 0.5at²
Acceleration of ball Y, x = 0.5a / 2 = ¼ a








Question 465: [Matter > Young modulus]
Lift is supported by two steel cables each of length 20 m.
Each of the cables consists of 100 parallel steel wires, each wire of cross-sectional area 3.2 × 10^–6 m². Young modulus of steel is 2.1 × 10¹¹ N m^–2.
Which distance does lift move downward when a man of mass 70 kg steps into it?
A 0.010 mm                B 0.020 mm                C 0.10 mm                  D 0.20 mm

Reference: Past Exam Paper – November 2013 Paper 13 Q22



Solution 465:
Answer: C.
Hooke’s law: F = ke

The lift is supported by 2 steel cables, each of which consists of 100 parallel steel wires. So, there is a total of 200 parallel steel wires in parallel.

For parallel spring,
Effective spring constant, k_eff = k₁ + k₂ + k₃ + …

First, we need to find the spring constant, k for 1 steel wire.

For 1 wire,
Young modulus, E = stress / strain = (F/A) / (e/L) = FL / Ae
Hooke’s law: F = ke
Young modulus, E = (ke)L / Ae = kL / A
Spring constant, k = EA / L = (2.1 × 10¹¹) (3.2 × 10^–6) / 20 = 33600 Nm^-1

Effective spring constant, k_eff = 200k
Hooke’s law: F = k_eff e
Mass of person = 70kg. Weight = mg = 700N           (take g = 10 ms^-2)
Extension, e = F / k_eff = 700 / (200 × 33600) = 0.00010m = 0.10 mm










Question 466: [Work, Energy, Power]
Diagram shows lift system in which the elevator (mass m₁) is partly counterbalanced by a heavy weight (mass m₂).

At what rate does motor provide energy to the system when elevator is rising at a steady speed v? (g = acceleration of free fall)
A ½ m₁v²                     B ½ (m₁ – m₂)v²                      C m₁gv            D (m₁ – m₂)gv

Reference: Past Exam Paper – November 2010 Paper 11 Q18



Solution 466:

Answer: D.

The rate at which the motor provides energy is the power of the motor.

The system works as such: the mass m₂ works downwards under gravity, causing the elevator of mass m₁ to move up. However, the motor also provides some energy in moving the elevator up.

The (gravitational) potential energy of the masses are changing continuously and at a steady rate, since they are moving at a steady speed.

GPE of elevator = m₁gh₁        where h₁ is its height above the ground

Rate of change of GPE of elevator = m₁gh₁ / t = m₁gv

GPE of heavy weight = m₂gh₂            where h₂ is its height above the ground

Rate of change of GPE of heavy weight = m₂gh₂ / t = m₂gv

Let the power of the motor be P.

m₁gv = m₂gv + P

Power of motor = (m₁ – m₂)gv

Choices A and B have units of energy, not power. If B was correct, then there would have no need for the heavy weight.